Another application of the monotone convergence theorem: $\textit{Beppo Levi's Lemma}$
Given a sequence of measurable functions $\{f_n\}_{n=1}^\infty$ where $f_n: E \subseteq X \rightarrow [0, \infty] $, $f_n \leq f_{n+1}$ and $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is a bounded sequence of (non-negative) real numbers, then
$f_n \longrightarrow f$ a.e., $f$ is finite a.e. and $\int_E f_n \longrightarrow \int_E f < \infty $
$\textit{Proof} $ : By the monotinicity of integration,
$f_n \leq f_{n+1} \Longrightarrow \int_E f_n \leq \int_E f_{n+1} $ and so $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is an increasing sequence in $[0,\infty)$ with a limit in $[0,\infty)$. Since $\{f_n\}$ is increasing, then for each fixed $x \in{E}$,
$$f(x) = \lim_{n \rightarrow \infty}f_n(x)$$
exists in $[0,\infty]$. Also, $f$ is measurable since it is the pointwise limit of a sequence of measurable functions. Now, the monotone convergence theorem applies and gives
$$\lim_{n \rightarrow \infty} \int_E f_n = \int_E f $$
Since $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is bounded, so is its limit, $ \int_E f$. Thus, $f$ must be finite a.e. To actually show $f$ is finite a.e. we can use Chebychev's inequality:
Define $E_n = \{x: f(x) \geq n\} = \{x: \frac{f(x)}{n} \geq 1 \}$, $n\in{\mathbb{N}}$. Then, by the linearity and monotinicity of integration,
$$m(E_n) = \int_{E_n}1 \leq \int_{E_n} \frac{f}{n} = \frac{1}{n} \int_{E_n} f \leq \frac{1}{n} \int_E f < \infty$$
Thus, $m(E_n) \overset{n \rightarrow \infty}{\longrightarrow} 0$ so in particular, $m(\{x: f(x) = \infty \}) = 0$ i.e. $f$ is finite a.e.
$\blacksquare$
Saturday, July 19, 2014
Monday, July 7, 2014
Riesz-Fischer
$L^p(X,\mu)$ is a complete normed space $\forall p \in{[1,\infty]}$ and any positive measure $\mu$
Case 1: $p=\infty$
The space $L^\infty$ consists of functions which are essentially bounded. Generally speaking, extended valued functions may take on the values of $\pm \infty $, in which case if we look at the sup norm of such functions, we may have $\sup_{x \in{X}}|f(x)|=\infty$. If we want to examine the bounds of a function almost everywhere, then we define the $\textit{essential sup}$ of $f$: $$\text{esssup}(f) = \inf\{M>0 : |f(x)| \leq M \quad a.e.\} = ||f||_\infty $$ This is a generalization of the usual sup norm and is the same as the sup norm for bounded functions such as $C[a,b]$. Some examples:
1. $f(x) = +\infty \chi_\mathbb{Q}$. Although $f$ takes on the value of $+\infty$ infinitely many times, it does so on a set of measure 0. Thus $||f||_\infty = 0$
2. $f(x) = -\log(x)\chi_{[0,1]}$. Then, $\int_0^1 \log^p(x) = \Gamma(p+1)$ and so $f\in{L^p[0,1]}$ for any finite $p$. However, $\forall M>0$, there exists an interval, namely $[0,e^{-M})$ which has positive measure, $e^{-M}$, where $f>M$. Thus, $f \not \in{L^\infty[0,1]}$
In general, we can make an inclusion relation between $L^p(E)$ and $L^q(E)$ only when $\mu(E)<\infty$.
$\textit{Lemma}: ||\cdot||_\infty$ is a norm.
(i) $f=0$ a.e. $\Longleftrightarrow$ $|f| \leq 0$ a.e. $\Longleftrightarrow$ $||f||_\infty = 0$.
(ii) Let $\alpha \in{\mathbb{R}}$ (or $\alpha\in{F}$...whatever underlying field of scalars our functions map to). Then, $||\alpha f||_\infty = \inf\{M \geq 0: |\alpha f(x)| \leq M \text{ a.e. x} \} = \inf\{M \geq 0: |\alpha| \cdot | f(x)| \leq M \text{ a.e. x} \} = |\alpha | \cdot \inf\{M \geq 0: | f(x)| \leq M \text{ a.e. x} \} = |\alpha| \cdot ||f||_\infty$
(iii) $||f+g||_\infty = \inf\{M \geq 0 : |f(x)+g(x)| \leq M \text{ a.e. x} \} \leq \inf\{M_1 \geq 0 : |f(x)| \leq M_1 \text{ a.e. x} \} + \inf\{M_2 \geq 0 : |g(x)| \leq M_2 \text{ a.e. x} \} = ||f||_\infty + ||g||_\infty $
We now wish to show that $L^\infty(E \subseteq X,\mu) = \{f: E\in{\mathcal{M}} \longrightarrow [-\infty,\infty]: \quad ||f||_\infty < \infty \}$ together with the norm $|| \cdot ||_\infty$ is complete: If $||f_n-f_m||_\infty \longrightarrow 0$ as $n,m \longrightarrow \infty$ then $||f_n-f||_\infty \longrightarrow 0$ where $f\in{L^\infty}$
$\textit{proof}$ :
Let $\{f_n\}$ be a Cauchy sequence with respect to $||\cdot||_\infty$. Define $$A_n = \{x \in{E} : |f_n(x)| > ||f_n||_\infty \} $$ $$B_{mn} = \{x\in{E}: |f_n(x)-f_m(x)| > ||f_n-f_m||_\infty \}$$ By definition, $\mu(A_n) = \mu(B_{mn}) = 0$, so $\mu(A_n \cup B_{mn}) \leq \mu(A_n) + \mu(B_{mn}) = 0+0 = 0$.
For any fixed $ x \in{(A_n \cup B_{mn})^c}$, $\{|f_n(x)|\}_{n=1}^\infty $ and $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ are bounded sequences of non-negative real numbers. $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ is a Cauchy sequence of real numbers and since $(\mathbb{R},| \cdot|)$ is complete, this sequence is convergent. We can see this as follows: On $(A_n \cup B_{mn})^c$, the $L^\infty$ norm is the same as the sup norm $$|f_n(x) - f_m(x)| \leq ||f_n - f_m||_\infty = \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f_m(x)| \longrightarrow 0 \text{ as } n,m \longrightarrow \infty$$ Convergence in the sup norm is equivalent to uniform convergence. Letting $m \longrightarrow \infty$, $$|f_n(x)-f(x)| \leq \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f(x)| \longrightarrow 0 \text{ as } n \longrightarrow \infty$$ So, $f_n(x) \longrightarrow f(x)$ uniformly $\forall x\in{(A_n \cup B_{mn})^c}$ thus forcing $f$ to be bounded a.e.
Case 2: $p=1$
Consider the following examples:
1. $f_n = \frac{1}{n}$. Then $f_n \longrightarrow 0$ (in fact, uniformly) on all of $\mathbb{R}$, yet $\int_\mathbb{R} f_n = \infty$ $\forall n \in{\mathbb{N}}$ while $\int_\mathbb{R}0 = 0$. This shows us that pointwise (or even uniform) convergence does not imply convergence in the $L^1$ norm.
2. Let $f_1 = \chi_{[0,1]}$, $f_2 = \chi_{[0,\frac{1}{2}]}$, $f_3 = \chi_{[\frac{1}{2},1]}$, $f_4 = \chi_{[0,\frac{1}{3}]}$, $f_5 = \chi_{[\frac{1}{3},\frac{2}{3}]}$, $f_6 = \chi_{[\frac{2}{3},1]}$, $f_7 = \chi_{[0,\frac{1}{4}]}$, ...
So, $||f_n||_1 \longrightarrow 0$ yet $\{f_n\}$ does not have a pointwise limit since the $f_n$ continue to "sweep" back from 0 to 1. Indeed, $\forall x\in{[0,1]}$, $\exists N_x \in{\mathbb{N}}$ s.t. $f_N(x) = 1$.
$\{f_n\}$ still manages to be Cauchy in $L^1$ since, for $n,m \in{\mathbb{N}}$, $||f_n-f_m||_1 \leq \frac{1}{n} + \frac{1}{m} \longrightarrow 0$ as $n,m \longrightarrow \infty$
Despite this, we can still manage to find a subsequence $\{f_{n_k} \}$ of $\{f_n\}$ that converges pointwise a.e. to $f=0$. For instance, the subsequence $f_1, f_2,f_4,f_7,...$ is s.t. $f_{n_k}= \chi_{[0,\frac{1}{k}]} \longrightarrow 0$ a.e. For an even faster converging subsequence, choose $f_2,f_7,...$. In this case, $||f_{n_{k+1}}-f_{n_k}||_1 = \int |\chi_{[0,\frac{1}{2^k}]}-\chi_{[0,\frac{1}{2^{k-1}}]}| = \int \chi_{[0,\frac{1}{2^k}]} = \frac{1}{2^k}$. These types of subsequences which converge pointwise as well as in the $L^1$ norm are called $\textit{rapidly Cauchy}$ subsequences.
$\textit{proof}$
Let $\{f_n\}_{n=1}^\infty$ be Cauchy in $L^1$. We wish to show that $||f_n-f||_1 \longrightarrow 0$ and that $f\in{L^1}$. Choose a subsequence $\{f_{n_k}\}$ that is "rapidly Cauchy" i.e. $$||f_{n_{k+1}} - f_{n_k}||_1 \leq \frac{1}{2^k} \quad \forall k\in{\mathbb{N}}$$ Such a subsequence exists from assumption that $\{f_n\}$ is Cauchy. By induction, for any $k \in{\mathbb{N}}$, we may choose an $n_k$ (now depends on $k$) sufficiently large s.t. the above inequality holds. Since $f_{n_k} $ are integrable, they are measurable. If we can show that $|f_{n_k}-f|$ is dominated by an $L^1$ function and $f_{n_k}-f \longrightarrow 0$ a.e. then the dominated convergence theorem will imply $||f_{n_k}-f||_1 \longrightarrow 0$. Since $\{f_n\}$ is assumed to be Cauchy, $||f_{n_k}-f_n|| \longrightarrow 0$. An "$\frac{\epsilon}{2}$" argument and the triangle inequality of the $L^1$ norm give $$||f_n-f||_1 = ||f_n-f_{n_k}+f_{n_k}-f||_1 \leq ||f_n-f_{n_k}||_1 + ||f_{n_k}-f||_1 \overset{n\rightarrow \infty}{\longrightarrow} 0$$ The key property of the extracted subsequence $\{f_{n_k}\}$ suggests that we should define $f$ in terms of a telescoping series. Let $$f(x) = f_{n_1}(x) + \sum_{k=1}^\infty(f_{n_{k+1}}(x)-f_{n_k}(x))$$ By the generalized triangle inequality and the continuity of the absolute value function, $$|f| = \left|f_{n_1} + \sum_{k=1}^\infty(f_{n_{k+1}}-f_{n_k})\right| \leq |f_{n_1}| + \left|\lim_{N \rightarrow \infty } \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right| = |f_{n_1}| + \lim_{N \rightarrow \infty }\left| \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right|$$ $$ \leq |f_{n_1}| + \lim_{N \rightarrow \infty } \sum_{k=1}^N\left|f_{n_{k+1}}-f_{n_k}\right| = |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ The above sequence of partial sums is non-negative and increasing, so apriori, the limit may be $\infty$. By the monotinicity and linearity of the Lebesgue integral, $$||f||_1 = \int|f| \leq \int|f_{n_1}| + \int \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}| = \int |f_{n_1}| + \sum_{k=1}^\infty \int |f_{n_{k+1}} - f_{n_k}|$$ where the justification for the interchange of countable sum and integral comes from noting that $0 \leq \sum_{k=1}^N |f_{n_{k+1}} - f_{n_k}| \nearrow \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}|$ and applying the monotone convergence theorem. $$=||f_{n_1}||_1 + \sum_{k=1}^\infty ||f_{n_{k+1}}-f_{n_k}||_1 \leq ||f_{n_1}||_1 + \sum_{k=1}^\infty \frac{1}{2^k} = ||f_{n_1}||_1 + 1 < \infty$$ So, $|f| \leq g\in{L^1} $ hence $ f\in{L^1}$. In particular, the series defining $f$ converges a.e. and since the partial sums are exactly $f_{n_k}$: $$f(x) = f_{n_1}(x) + \lim_{N \rightarrow \infty} \sum_{k=1}^N (f_{n_{k+1}}(x)-f_{n_k}(x)) = \lim_{N \rightarrow \infty}f_{n_N}(x) \quad \text{a.e. } x$$ So, $\{f_{n_k}-f\}$ is a sequence of $L^1$ (hence measurable) functions and $f_{n_k}-f \longrightarrow 0$ a.e. Also, $$|f_{n_N}| = |f_{n_1}| + \sum_{k=1}^N|f_{n_{k+1}}-f_{n_k}| \leq |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ By the triangle inequality, $|f_{n_k}-f| \leq |f|+|f_{n_k}| \leq g+g=2g \in{L^1}$. Applying the dominated convergence theorem gives $||f_{n_k}-f||_1 \longrightarrow 0$ The case for $1
The space $L^\infty$ consists of functions which are essentially bounded. Generally speaking, extended valued functions may take on the values of $\pm \infty $, in which case if we look at the sup norm of such functions, we may have $\sup_{x \in{X}}|f(x)|=\infty$. If we want to examine the bounds of a function almost everywhere, then we define the $\textit{essential sup}$ of $f$: $$\text{esssup}(f) = \inf\{M>0 : |f(x)| \leq M \quad a.e.\} = ||f||_\infty $$ This is a generalization of the usual sup norm and is the same as the sup norm for bounded functions such as $C[a,b]$. Some examples:
1. $f(x) = +\infty \chi_\mathbb{Q}$. Although $f$ takes on the value of $+\infty$ infinitely many times, it does so on a set of measure 0. Thus $||f||_\infty = 0$
2. $f(x) = -\log(x)\chi_{[0,1]}$. Then, $\int_0^1 \log^p(x) = \Gamma(p+1)$ and so $f\in{L^p[0,1]}$ for any finite $p$. However, $\forall M>0$, there exists an interval, namely $[0,e^{-M})$ which has positive measure, $e^{-M}$, where $f>M$. Thus, $f \not \in{L^\infty[0,1]}$
In general, we can make an inclusion relation between $L^p(E)$ and $L^q(E)$ only when $\mu(E)<\infty$.
$\textit{Lemma}: ||\cdot||_\infty$ is a norm.
(i) $f=0$ a.e. $\Longleftrightarrow$ $|f| \leq 0$ a.e. $\Longleftrightarrow$ $||f||_\infty = 0$.
(ii) Let $\alpha \in{\mathbb{R}}$ (or $\alpha\in{F}$...whatever underlying field of scalars our functions map to). Then, $||\alpha f||_\infty = \inf\{M \geq 0: |\alpha f(x)| \leq M \text{ a.e. x} \} = \inf\{M \geq 0: |\alpha| \cdot | f(x)| \leq M \text{ a.e. x} \} = |\alpha | \cdot \inf\{M \geq 0: | f(x)| \leq M \text{ a.e. x} \} = |\alpha| \cdot ||f||_\infty$
(iii) $||f+g||_\infty = \inf\{M \geq 0 : |f(x)+g(x)| \leq M \text{ a.e. x} \} \leq \inf\{M_1 \geq 0 : |f(x)| \leq M_1 \text{ a.e. x} \} + \inf\{M_2 \geq 0 : |g(x)| \leq M_2 \text{ a.e. x} \} = ||f||_\infty + ||g||_\infty $
We now wish to show that $L^\infty(E \subseteq X,\mu) = \{f: E\in{\mathcal{M}} \longrightarrow [-\infty,\infty]: \quad ||f||_\infty < \infty \}$ together with the norm $|| \cdot ||_\infty$ is complete: If $||f_n-f_m||_\infty \longrightarrow 0$ as $n,m \longrightarrow \infty$ then $||f_n-f||_\infty \longrightarrow 0$ where $f\in{L^\infty}$
$\textit{proof}$ :
Let $\{f_n\}$ be a Cauchy sequence with respect to $||\cdot||_\infty$. Define $$A_n = \{x \in{E} : |f_n(x)| > ||f_n||_\infty \} $$ $$B_{mn} = \{x\in{E}: |f_n(x)-f_m(x)| > ||f_n-f_m||_\infty \}$$ By definition, $\mu(A_n) = \mu(B_{mn}) = 0$, so $\mu(A_n \cup B_{mn}) \leq \mu(A_n) + \mu(B_{mn}) = 0+0 = 0$.
For any fixed $ x \in{(A_n \cup B_{mn})^c}$, $\{|f_n(x)|\}_{n=1}^\infty $ and $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ are bounded sequences of non-negative real numbers. $\{|f_n(x)-f_m(x)|\}_{n,m=1}^\infty$ is a Cauchy sequence of real numbers and since $(\mathbb{R},| \cdot|)$ is complete, this sequence is convergent. We can see this as follows: On $(A_n \cup B_{mn})^c$, the $L^\infty$ norm is the same as the sup norm $$|f_n(x) - f_m(x)| \leq ||f_n - f_m||_\infty = \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f_m(x)| \longrightarrow 0 \text{ as } n,m \longrightarrow \infty$$ Convergence in the sup norm is equivalent to uniform convergence. Letting $m \longrightarrow \infty$, $$|f_n(x)-f(x)| \leq \sup_{x\in{(A_n\cup B{mn})^c}}|f_n(x)-f(x)| \longrightarrow 0 \text{ as } n \longrightarrow \infty$$ So, $f_n(x) \longrightarrow f(x)$ uniformly $\forall x\in{(A_n \cup B_{mn})^c}$ thus forcing $f$ to be bounded a.e.
Case 2: $p=1$
Consider the following examples:
1. $f_n = \frac{1}{n}$. Then $f_n \longrightarrow 0$ (in fact, uniformly) on all of $\mathbb{R}$, yet $\int_\mathbb{R} f_n = \infty$ $\forall n \in{\mathbb{N}}$ while $\int_\mathbb{R}0 = 0$. This shows us that pointwise (or even uniform) convergence does not imply convergence in the $L^1$ norm.
2. Let $f_1 = \chi_{[0,1]}$, $f_2 = \chi_{[0,\frac{1}{2}]}$, $f_3 = \chi_{[\frac{1}{2},1]}$, $f_4 = \chi_{[0,\frac{1}{3}]}$, $f_5 = \chi_{[\frac{1}{3},\frac{2}{3}]}$, $f_6 = \chi_{[\frac{2}{3},1]}$, $f_7 = \chi_{[0,\frac{1}{4}]}$, ...
So, $||f_n||_1 \longrightarrow 0$ yet $\{f_n\}$ does not have a pointwise limit since the $f_n$ continue to "sweep" back from 0 to 1. Indeed, $\forall x\in{[0,1]}$, $\exists N_x \in{\mathbb{N}}$ s.t. $f_N(x) = 1$.
$\{f_n\}$ still manages to be Cauchy in $L^1$ since, for $n,m \in{\mathbb{N}}$, $||f_n-f_m||_1 \leq \frac{1}{n} + \frac{1}{m} \longrightarrow 0$ as $n,m \longrightarrow \infty$
Despite this, we can still manage to find a subsequence $\{f_{n_k} \}$ of $\{f_n\}$ that converges pointwise a.e. to $f=0$. For instance, the subsequence $f_1, f_2,f_4,f_7,...$ is s.t. $f_{n_k}= \chi_{[0,\frac{1}{k}]} \longrightarrow 0$ a.e. For an even faster converging subsequence, choose $f_2,f_7,...$. In this case, $||f_{n_{k+1}}-f_{n_k}||_1 = \int |\chi_{[0,\frac{1}{2^k}]}-\chi_{[0,\frac{1}{2^{k-1}}]}| = \int \chi_{[0,\frac{1}{2^k}]} = \frac{1}{2^k}$. These types of subsequences which converge pointwise as well as in the $L^1$ norm are called $\textit{rapidly Cauchy}$ subsequences.
$\textit{proof}$
Let $\{f_n\}_{n=1}^\infty$ be Cauchy in $L^1$. We wish to show that $||f_n-f||_1 \longrightarrow 0$ and that $f\in{L^1}$. Choose a subsequence $\{f_{n_k}\}$ that is "rapidly Cauchy" i.e. $$||f_{n_{k+1}} - f_{n_k}||_1 \leq \frac{1}{2^k} \quad \forall k\in{\mathbb{N}}$$ Such a subsequence exists from assumption that $\{f_n\}$ is Cauchy. By induction, for any $k \in{\mathbb{N}}$, we may choose an $n_k$ (now depends on $k$) sufficiently large s.t. the above inequality holds. Since $f_{n_k} $ are integrable, they are measurable. If we can show that $|f_{n_k}-f|$ is dominated by an $L^1$ function and $f_{n_k}-f \longrightarrow 0$ a.e. then the dominated convergence theorem will imply $||f_{n_k}-f||_1 \longrightarrow 0$. Since $\{f_n\}$ is assumed to be Cauchy, $||f_{n_k}-f_n|| \longrightarrow 0$. An "$\frac{\epsilon}{2}$" argument and the triangle inequality of the $L^1$ norm give $$||f_n-f||_1 = ||f_n-f_{n_k}+f_{n_k}-f||_1 \leq ||f_n-f_{n_k}||_1 + ||f_{n_k}-f||_1 \overset{n\rightarrow \infty}{\longrightarrow} 0$$ The key property of the extracted subsequence $\{f_{n_k}\}$ suggests that we should define $f$ in terms of a telescoping series. Let $$f(x) = f_{n_1}(x) + \sum_{k=1}^\infty(f_{n_{k+1}}(x)-f_{n_k}(x))$$ By the generalized triangle inequality and the continuity of the absolute value function, $$|f| = \left|f_{n_1} + \sum_{k=1}^\infty(f_{n_{k+1}}-f_{n_k})\right| \leq |f_{n_1}| + \left|\lim_{N \rightarrow \infty } \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right| = |f_{n_1}| + \lim_{N \rightarrow \infty }\left| \sum_{k=1}^N(f_{n_{k+1}}-f_{n_k})\right|$$ $$ \leq |f_{n_1}| + \lim_{N \rightarrow \infty } \sum_{k=1}^N\left|f_{n_{k+1}}-f_{n_k}\right| = |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ The above sequence of partial sums is non-negative and increasing, so apriori, the limit may be $\infty$. By the monotinicity and linearity of the Lebesgue integral, $$||f||_1 = \int|f| \leq \int|f_{n_1}| + \int \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}| = \int |f_{n_1}| + \sum_{k=1}^\infty \int |f_{n_{k+1}} - f_{n_k}|$$ where the justification for the interchange of countable sum and integral comes from noting that $0 \leq \sum_{k=1}^N |f_{n_{k+1}} - f_{n_k}| \nearrow \sum_{k=1}^\infty |f_{n_{k+1}} - f_{n_k}|$ and applying the monotone convergence theorem. $$=||f_{n_1}||_1 + \sum_{k=1}^\infty ||f_{n_{k+1}}-f_{n_k}||_1 \leq ||f_{n_1}||_1 + \sum_{k=1}^\infty \frac{1}{2^k} = ||f_{n_1}||_1 + 1 < \infty$$ So, $|f| \leq g\in{L^1} $ hence $ f\in{L^1}$. In particular, the series defining $f$ converges a.e. and since the partial sums are exactly $f_{n_k}$: $$f(x) = f_{n_1}(x) + \lim_{N \rightarrow \infty} \sum_{k=1}^N (f_{n_{k+1}}(x)-f_{n_k}(x)) = \lim_{N \rightarrow \infty}f_{n_N}(x) \quad \text{a.e. } x$$ So, $\{f_{n_k}-f\}$ is a sequence of $L^1$ (hence measurable) functions and $f_{n_k}-f \longrightarrow 0$ a.e. Also, $$|f_{n_N}| = |f_{n_1}| + \sum_{k=1}^N|f_{n_{k+1}}-f_{n_k}| \leq |f_{n_1}| + \sum_{k=1}^\infty|f_{n_{k+1}}-f_{n_k}|=g$$ By the triangle inequality, $|f_{n_k}-f| \leq |f|+|f_{n_k}| \leq g+g=2g \in{L^1}$. Applying the dominated convergence theorem gives $||f_{n_k}-f||_1 \longrightarrow 0$ The case for $1
Monday, June 23, 2014
Stein 1.25
An alternative definition of measurability is as follows: $E$ is measurable if $\forall \epsilon > 0$, $ \exists$ closed set $F \subseteq E$ s.t. $m_*(E-F) \leq \epsilon$ $\star$. Show that this definition is equivalent (in the sense that each definition implies the other) to the definition given earlier in the text:
$E$ is meas. if $\forall \epsilon > 0$, $\exists \mathcal{O}_{open} \supseteq E$ s.t. $m_*(\mathcal{O}-E) \leq \epsilon$ $\star\star$
$\textit{Proof}$ : Let $\mathcal{M}$ be the set of Lebesgue measurable sets according to the definition $\star\star$ given in the text. Let $\mathcal{M}'$ be the collection of measurable sets given by $\star$. Shwoing that the two definitions are equivalent amounts to showing that $\mathcal{M} \subseteq \mathcal{M}'$ and $\mathcal{M}' \subseteq \mathcal{M}$.
"$\subseteq$" Let $E\in{\mathcal{M}}$. Then, since $\mathcal{M}$ is a $\sigma-$algebra, $E^c \in{\mathcal{M}}$. So, $\forall \epsilon >0$, $\exists \mathcal{O}_{open} \supseteq E^c$ s.t. $m_*(\mathcal{O}-E^c) \leq \epsilon$. $\mathcal{O}^c$ is closed since $\mathcal{O}$ is open. So, $\mathcal{O} \supseteq E^c \Longrightarrow \mathcal{O}^c \subseteq E$. If we can show that $E-\mathcal{O}^c \subseteq \mathcal{O}-E^c$ then we are done since the monotinicity of outer measure then implies that $m_*(E-\mathcal{O}^c) \leq m_*(\mathcal{O}-E^c) \leq \epsilon$.
Let $x\in{(E-\mathcal{O}^c)} \Longleftrightarrow x\in{E}$ and $x\not \in{\mathcal{O}^c} \Longleftrightarrow x\in{E}$ and $x\in{\mathcal{O}}$ $ \Longleftrightarrow x\in{\mathcal{O}}$ and $x\not \in{E}^c \Longleftrightarrow x\in{(\mathcal{O}-E^c)}$
Thus, $ \mathcal{M} \subseteq \mathcal{M}'$
"$\supseteq$" Let $E \in{\mathcal{M}'} \Longrightarrow \forall \epsilon >0 \quad \exists F_{closed} \subseteq E$ s.t. $m_*(E-F) \leq \epsilon$. Now, $F \subseteq E \Longrightarrow E^c \subseteq F_{open}^c$. If we can show that $F^c - E^c \subseteq E-F$ then the monotinicity of outer measure again implies $m_*(F^c-E^c) \leq m_*(E-F) \leq \epsilon$.
Let $x\in{(F^c-E^c)} \Longleftrightarrow x\not \in{F}$ and $x\not \in{E}^c \Longleftrightarrow x\in{E}$ and $x\not \in{F} \Longleftrightarrow x\in{(E-F)}$. By definition $\star$ this implies that $E^c \in{\mathcal{M}}$ which implies $E \in{\mathcal{M}}$.
Thus, $\mathcal{M} \supseteq \mathcal{M}'$.
$\therefore \mathcal{M}=\mathcal{M}'$ $\blacksquare$
$E$ is meas. if $\forall \epsilon > 0$, $\exists \mathcal{O}_{open} \supseteq E$ s.t. $m_*(\mathcal{O}-E) \leq \epsilon$ $\star\star$
$\textit{Proof}$ : Let $\mathcal{M}$ be the set of Lebesgue measurable sets according to the definition $\star\star$ given in the text. Let $\mathcal{M}'$ be the collection of measurable sets given by $\star$. Shwoing that the two definitions are equivalent amounts to showing that $\mathcal{M} \subseteq \mathcal{M}'$ and $\mathcal{M}' \subseteq \mathcal{M}$.
"$\subseteq$" Let $E\in{\mathcal{M}}$. Then, since $\mathcal{M}$ is a $\sigma-$algebra, $E^c \in{\mathcal{M}}$. So, $\forall \epsilon >0$, $\exists \mathcal{O}_{open} \supseteq E^c$ s.t. $m_*(\mathcal{O}-E^c) \leq \epsilon$. $\mathcal{O}^c$ is closed since $\mathcal{O}$ is open. So, $\mathcal{O} \supseteq E^c \Longrightarrow \mathcal{O}^c \subseteq E$. If we can show that $E-\mathcal{O}^c \subseteq \mathcal{O}-E^c$ then we are done since the monotinicity of outer measure then implies that $m_*(E-\mathcal{O}^c) \leq m_*(\mathcal{O}-E^c) \leq \epsilon$.
Let $x\in{(E-\mathcal{O}^c)} \Longleftrightarrow x\in{E}$ and $x\not \in{\mathcal{O}^c} \Longleftrightarrow x\in{E}$ and $x\in{\mathcal{O}}$ $ \Longleftrightarrow x\in{\mathcal{O}}$ and $x\not \in{E}^c \Longleftrightarrow x\in{(\mathcal{O}-E^c)}$
Thus, $ \mathcal{M} \subseteq \mathcal{M}'$
"$\supseteq$" Let $E \in{\mathcal{M}'} \Longrightarrow \forall \epsilon >0 \quad \exists F_{closed} \subseteq E$ s.t. $m_*(E-F) \leq \epsilon$. Now, $F \subseteq E \Longrightarrow E^c \subseteq F_{open}^c$. If we can show that $F^c - E^c \subseteq E-F$ then the monotinicity of outer measure again implies $m_*(F^c-E^c) \leq m_*(E-F) \leq \epsilon$.
Let $x\in{(F^c-E^c)} \Longleftrightarrow x\not \in{F}$ and $x\not \in{E}^c \Longleftrightarrow x\in{E}$ and $x\not \in{F} \Longleftrightarrow x\in{(E-F)}$. By definition $\star$ this implies that $E^c \in{\mathcal{M}}$ which implies $E \in{\mathcal{M}}$.
Thus, $\mathcal{M} \supseteq \mathcal{M}'$.
$\therefore \mathcal{M}=\mathcal{M}'$ $\blacksquare$
Saturday, June 21, 2014
Stein 1.22
Prove that there is no continuous function $f$ on $\mathbb{R}$ s.t. $f= \chi_{[0,1]}$ a.e.
$\textit{Proof} :$
Suppose to the contrary that $f\in{C(\mathbb{R})}$ and $f = \chi_{[0,1]}$ a.e. With $f$ continuous, $\forall \epsilon > 0 $, $\exists \delta = \delta(x,\epsilon) > 0 $ s.t. $|x-z| \leq \delta \Longrightarrow |f(x) - f(z)| \leq \epsilon$. Fix $\epsilon = \frac{1}{2}$, let $x\in{[0,1]}$ and $z <0$. Then, $$\chi_{[0,1]}(x) - \chi_{[0,1]}(z) = 1-0 = 1 \quad \forall x\in{[0,1]}, z<0 $$ and so $$f(x) - f(z) = 1 \quad \text{a.e. } x\in{[0,1]}, z<0 $$ Let $\delta > 0$ and let $x\in{[0,1]},z<0$ s.t. $0 < m(|x-z|)=x-z \leq \delta$.
However, $f(x)-f(z) = 1$ a.e. In particular, $f(x_0)-f(z_0) = 1 $ for some $x_0\in{[0,1]}$, $z_0 < 0$, $|x_0-z_0| \leq \delta$...contradiction. (If we could not find such an $x_0$ and $z_0$ as above, that would contradict the assumption that $f = \chi_{[0,1]}$ a.e.) $\blacksquare$
$\textit{Proof} :$
Suppose to the contrary that $f\in{C(\mathbb{R})}$ and $f = \chi_{[0,1]}$ a.e. With $f$ continuous, $\forall \epsilon > 0 $, $\exists \delta = \delta(x,\epsilon) > 0 $ s.t. $|x-z| \leq \delta \Longrightarrow |f(x) - f(z)| \leq \epsilon$. Fix $\epsilon = \frac{1}{2}$, let $x\in{[0,1]}$ and $z <0$. Then, $$\chi_{[0,1]}(x) - \chi_{[0,1]}(z) = 1-0 = 1 \quad \forall x\in{[0,1]}, z<0 $$ and so $$f(x) - f(z) = 1 \quad \text{a.e. } x\in{[0,1]}, z<0 $$ Let $\delta > 0$ and let $x\in{[0,1]},z<0$ s.t. $0 < m(|x-z|)=x-z \leq \delta$.
However, $f(x)-f(z) = 1$ a.e. In particular, $f(x_0)-f(z_0) = 1 $ for some $x_0\in{[0,1]}$, $z_0 < 0$, $|x_0-z_0| \leq \delta$...contradiction. (If we could not find such an $x_0$ and $z_0$ as above, that would contradict the assumption that $f = \chi_{[0,1]}$ a.e.) $\blacksquare$
Stein 1.21
Show that there exists a continuous function $F$ that maps a measurable set to a non-measurable set.
$\textit{Example:}$ Consider $F$ to be the Cantor-Lebesgue function $$F: \mathcal{C} \longrightarrow [0,1]$$ which is continuous (in fact, since $\mathcal{C}$ is compact, $F$ is uniformly continuous). Exercise 32(b) establishes that given any subset of $\mathbb{R}$ with positive outer measure, we can embed a non-measurable subset in it. So, let $\mathcal{N} \subset [0,1]$ be non-measurable. Then, $F^{-1}(N) \subset \mathcal{C}$. Since $m(\mathcal{C})=0$ then $m(F^{-1}(N)) = 0$ and so $F^{-1}(N)$ is measurable. Of course, the restriction of a continuous function is still continuous. So, $$F|_{F^{-1}(N)}$$ is a continuous function that maps a measurable set to a non-measurable set. This gives an example showing that continuity is not a sufficiently strong enough condition to preserve measurability.
For the converse, consider $$\phi: \mathcal{N} \longrightarrow \{0\} \quad \text{where} \quad \phi = 0$$ Then, $\phi$ is trivially continuous and maps a non-measurable set to a measurable set.
One such condition on a function to preserve measurability is absolute continuity. Problem 3.19 establishes that if $f\in{AC(\mathbb{R})}$ then
(a) $f$ maps sets of measure 0 to sets of measure 0
(b) $f$ maps measurable sets to measurable sets
Part (a) is also known as the Lusin N property.
$\textit{Example:}$ Consider $F$ to be the Cantor-Lebesgue function $$F: \mathcal{C} \longrightarrow [0,1]$$ which is continuous (in fact, since $\mathcal{C}$ is compact, $F$ is uniformly continuous). Exercise 32(b) establishes that given any subset of $\mathbb{R}$ with positive outer measure, we can embed a non-measurable subset in it. So, let $\mathcal{N} \subset [0,1]$ be non-measurable. Then, $F^{-1}(N) \subset \mathcal{C}$. Since $m(\mathcal{C})=0$ then $m(F^{-1}(N)) = 0$ and so $F^{-1}(N)$ is measurable. Of course, the restriction of a continuous function is still continuous. So, $$F|_{F^{-1}(N)}$$ is a continuous function that maps a measurable set to a non-measurable set. This gives an example showing that continuity is not a sufficiently strong enough condition to preserve measurability.
For the converse, consider $$\phi: \mathcal{N} \longrightarrow \{0\} \quad \text{where} \quad \phi = 0$$ Then, $\phi$ is trivially continuous and maps a non-measurable set to a measurable set.
One such condition on a function to preserve measurability is absolute continuity. Problem 3.19 establishes that if $f\in{AC(\mathbb{R})}$ then
(a) $f$ maps sets of measure 0 to sets of measure 0
(b) $f$ maps measurable sets to measurable sets
Part (a) is also known as the Lusin N property.
Stein 1.16: The Boel Cantelli Lemma
$ \textit{The Borel-Cantelli Lemma} $
Let $(X,\mathcal{M},\mu)$ be a measure space. Given $\{E_k\}_{k=1}^\infty$ is a sequence of $\mu-$measurable subsets of $X$ and that $\sum_{k=1}^\infty \mu(E_k) < \infty$. Then, $\limsup E_k$ is also $\mu-$ measurable with measure 0.
$\textit{Preliminaries:}$ The $\limsup$ of a sequence of sets can be defined in several ways. The initial definition given in the book is $$\limsup E_k = \{x \in{X} : x\in{E_k} \quad \text{for infinitely many } k\}$$ Similarly, $$\liminf E_k = \{x \in{X} : x\in{E_k} \quad \text{for all but finitely many } k\}$$ The usual $\limsup$ of a sequence of real numbers $\{a_n\}$ gives us, in some sense, the number which is the largest accumulation point of $\{a_n\}$. Here, "largest" refers to comparison of real numbers with the ordering relation $" \leq "$. When looking at the generalized case when terms in our sequence are sets, the "largest accumulation set" is determined by the ordering relation on sets $" \subseteq "$. Equivalently, $$\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right) \quad \text{and} \quad \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$$ $\textit{Example 1}$: Let $X=\{0,1\}$ and let $E_k = \{\{-2\},\{4\},\{-1\}, \{0\},\{1\},\{0\},\{1\},...\}$.
So, $E_1 \cup E_2 \cup \cdots = \{0\} \cup \{1\} = \{0,1\}$. So $\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)= \bigcap_{k=1}^\infty \{0,1\} = \{0,1\}$...the set of $x\in{X}$ s.t. $x$ is in infinitely many $E_k$. On the other hand, $E_1 \cap E_2 \cap \cdots = \emptyset$ and so $\liminf E_k = \emptyset$...the only subset of $X$ that is in all but finitely many $E_k$. Note that in general, $\liminf E_k \subseteq \limsup E_k $: If $x$ is in all but finitely many $E_k$, then $x$ is in infinitely many $E_k$. However, if we consider our counting index to be $\mathbb{Z}$, $\{E_k\}_{k=-\infty}^{k+ \infty} $, then if, say, $x$ is in $E_k$ for all positive $k$, then $x$ is in infinitely many $E_k$, so $x\in{\limsup E_k }$. However, $x$ is still not in infinitely many $E_k$. So $x \not \in{\liminf E_k}$.
$\textit{Proof}$ :
First note that $\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)$ and $ \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$ are just combinations of countable unions and intersections of measurable $E_k$, so are measurable. $\sum_{k=1}^\infty \mu(E_k) < \infty \Longrightarrow \mu(E_k) \longrightarrow 0$ i.e. $\forall \epsilon > 0$, $\exists N\in{\mathbb{N}}$ s.t. as soon as $k \geq N$ then $\mu(E_k) \leq \dfrac{\epsilon}{2^k}$. Since $\bigcap_{N=1}^\infty(\bigcup_{k \geq N} E_k ) \subseteq \bigcup_{k \geq N} E_k $ then, since $\mu$ is a measure, it is monotonic and sub-additive: $$\mu(\limsup E_k) = \mu \left(\bigcap_{N=1}^\infty\left(\bigcup_{k \geq N} E_k \right) \right) \leq \mu \left( \bigcup_{k \geq N} E_k \right) \leq \sum_{k=N}^\infty \mu(E_k) $$ $$\leq \sum_{j=1}^\infty \mu(E_k) \leq \sum_{j=1}^\infty \frac{\epsilon}{2^k} = \epsilon \sum_{j=1}^\infty \frac{1}{2^k} = \epsilon$$ Since $\epsilon > 0 $ was chosen arbitrarily, we get $\mu(\limsup E_k) = 0$ $\blacksquare$
Let $(X,\mathcal{M},\mu)$ be a measure space. Given $\{E_k\}_{k=1}^\infty$ is a sequence of $\mu-$measurable subsets of $X$ and that $\sum_{k=1}^\infty \mu(E_k) < \infty$. Then, $\limsup E_k$ is also $\mu-$ measurable with measure 0.
$\textit{Preliminaries:}$ The $\limsup$ of a sequence of sets can be defined in several ways. The initial definition given in the book is $$\limsup E_k = \{x \in{X} : x\in{E_k} \quad \text{for infinitely many } k\}$$ Similarly, $$\liminf E_k = \{x \in{X} : x\in{E_k} \quad \text{for all but finitely many } k\}$$ The usual $\limsup$ of a sequence of real numbers $\{a_n\}$ gives us, in some sense, the number which is the largest accumulation point of $\{a_n\}$. Here, "largest" refers to comparison of real numbers with the ordering relation $" \leq "$. When looking at the generalized case when terms in our sequence are sets, the "largest accumulation set" is determined by the ordering relation on sets $" \subseteq "$. Equivalently, $$\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right) \quad \text{and} \quad \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$$ $\textit{Example 1}$: Let $X=\{0,1\}$ and let $E_k = \{\{-2\},\{4\},\{-1\}, \{0\},\{1\},\{0\},\{1\},...\}$.
So, $E_1 \cup E_2 \cup \cdots = \{0\} \cup \{1\} = \{0,1\}$. So $\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)= \bigcap_{k=1}^\infty \{0,1\} = \{0,1\}$...the set of $x\in{X}$ s.t. $x$ is in infinitely many $E_k$. On the other hand, $E_1 \cap E_2 \cap \cdots = \emptyset$ and so $\liminf E_k = \emptyset$...the only subset of $X$ that is in all but finitely many $E_k$. Note that in general, $\liminf E_k \subseteq \limsup E_k $: If $x$ is in all but finitely many $E_k$, then $x$ is in infinitely many $E_k$. However, if we consider our counting index to be $\mathbb{Z}$, $\{E_k\}_{k=-\infty}^{k+ \infty} $, then if, say, $x$ is in $E_k$ for all positive $k$, then $x$ is in infinitely many $E_k$, so $x\in{\limsup E_k }$. However, $x$ is still not in infinitely many $E_k$. So $x \not \in{\liminf E_k}$.
$\textit{Proof}$ :
First note that $\limsup E_k=\bigcap_{k=1}^\infty\left(\bigcup_{n \geq k} E_k\right)$ and $ \liminf E_k = \bigcup_{k=1}^\infty \left(\bigcap_{n \geq k} E_k \right)$ are just combinations of countable unions and intersections of measurable $E_k$, so are measurable. $\sum_{k=1}^\infty \mu(E_k) < \infty \Longrightarrow \mu(E_k) \longrightarrow 0$ i.e. $\forall \epsilon > 0$, $\exists N\in{\mathbb{N}}$ s.t. as soon as $k \geq N$ then $\mu(E_k) \leq \dfrac{\epsilon}{2^k}$. Since $\bigcap_{N=1}^\infty(\bigcup_{k \geq N} E_k ) \subseteq \bigcup_{k \geq N} E_k $ then, since $\mu$ is a measure, it is monotonic and sub-additive: $$\mu(\limsup E_k) = \mu \left(\bigcap_{N=1}^\infty\left(\bigcup_{k \geq N} E_k \right) \right) \leq \mu \left( \bigcup_{k \geq N} E_k \right) \leq \sum_{k=N}^\infty \mu(E_k) $$ $$\leq \sum_{j=1}^\infty \mu(E_k) \leq \sum_{j=1}^\infty \frac{\epsilon}{2^k} = \epsilon \sum_{j=1}^\infty \frac{1}{2^k} = \epsilon$$ Since $\epsilon > 0 $ was chosen arbitrarily, we get $\mu(\limsup E_k) = 0$ $\blacksquare$
Thursday, June 19, 2014
Stein 1.7
$\textbf{Exercise 1.7}$ Let $\delta=(\delta_1\delta_2 \cdots \delta_d)$ be a $d-$tuple of positive real numbers. Let $E \in{\mathcal{M}}$ and define
$$\delta E = \{(\delta_1x_1,...,\delta_dx_d) : (x_1,...,x_d) \in{E} \}$$
So, $\delta E$ is a dilation of the set $E$ by an amount of $\delta_j$ along each $j$-axis of $\mathbb{R}^d$. (This notation suggests that we are using the standard basis for $\mathbb{R}^d$).
Show that $\delta E \in{\mathcal{M}}$ and that $m(\delta E) = \delta_1 \cdots \delta_d \cdot m(E)$
$\textit{Proof}: $
$E \in{\mathcal{M}}$ means that $\forall \epsilon >0$, $\exists \mathcal{O} \supseteq E$, $\mathcal{O}$ open where $m_*(\mathcal{O}-E) \leq \epsilon$. By definition, $m_*(\mathcal{O}-E)= \inf \sum_{j=1}^\infty |Q_j|$ where the infimum is taken over all countable coverings of closed cubes, $\mathcal{O}- E \subseteq \bigcup_{j=1}^\infty Q_j$. Since the infimum of a sequence of real numbers is, by definition, an accumulation point, this means that there exists a sufficiently close covering of $\mathcal{O}-E$ s.t. $\sum_{j=1}^\infty |Q_j| \leq \epsilon $. Since $E$ has a covering by cubes, then $\delta E$ has a covering by rectangles, which can be approximated by cubes. So, with $$\mathcal{O}-E \subseteq \bigcup_{j=1}^\infty Q_j \quad \text{then} \quad \delta(\mathcal{O}-E)=\delta \mathcal{O}-\delta E \subseteq \bigcup_{j=1}^\infty \delta Q_j$$ Using the monotinicity and sub-additivity of $m_*$ $$m_*(\delta \mathcal{O}-\delta E) \leq m_*\left( \bigcup_{j=1}^\infty \delta Q_j \right) \leq \sum_{j=1}^\infty m_*(\delta Q_j) = \sum_{j=1}^\infty |\delta Q_j|$$ Since the side lengths of cubes align with the respective coordinate axes in $\mathbb{R}^d$, the $\delta$-dilation of a cube is known: $|\delta Q_j|=|\delta_1 \cdots \delta_d| \cdot |Q_j|$. So, $$\cdots = \sum_{j=1}^\infty |\delta_1 \cdots \delta_d| \cdot | Q_j| = |\delta_1 \cdots \delta_d| \sum_{j=1}^\infty |Q_j| \leq \delta_1 \cdots \delta_d \cdot \epsilon $$ Thus, $\delta E \in{\mathcal{M}}$. Also, $$m(\delta E) = m_*(\delta E) = \inf \sum_{j=1}^\infty |\delta Q_j| = \delta_1 \cdots \delta_d \cdot \inf \sum_{j=1}^\infty |Q_j|=\delta_1 \cdots \delta_d \cdot m_*(E) = \delta_1 \cdots \delta_d \cdot m(E)$$
Show that $\delta E \in{\mathcal{M}}$ and that $m(\delta E) = \delta_1 \cdots \delta_d \cdot m(E)$
$\textit{Proof}: $
$E \in{\mathcal{M}}$ means that $\forall \epsilon >0$, $\exists \mathcal{O} \supseteq E$, $\mathcal{O}$ open where $m_*(\mathcal{O}-E) \leq \epsilon$. By definition, $m_*(\mathcal{O}-E)= \inf \sum_{j=1}^\infty |Q_j|$ where the infimum is taken over all countable coverings of closed cubes, $\mathcal{O}- E \subseteq \bigcup_{j=1}^\infty Q_j$. Since the infimum of a sequence of real numbers is, by definition, an accumulation point, this means that there exists a sufficiently close covering of $\mathcal{O}-E$ s.t. $\sum_{j=1}^\infty |Q_j| \leq \epsilon $. Since $E$ has a covering by cubes, then $\delta E$ has a covering by rectangles, which can be approximated by cubes. So, with $$\mathcal{O}-E \subseteq \bigcup_{j=1}^\infty Q_j \quad \text{then} \quad \delta(\mathcal{O}-E)=\delta \mathcal{O}-\delta E \subseteq \bigcup_{j=1}^\infty \delta Q_j$$ Using the monotinicity and sub-additivity of $m_*$ $$m_*(\delta \mathcal{O}-\delta E) \leq m_*\left( \bigcup_{j=1}^\infty \delta Q_j \right) \leq \sum_{j=1}^\infty m_*(\delta Q_j) = \sum_{j=1}^\infty |\delta Q_j|$$ Since the side lengths of cubes align with the respective coordinate axes in $\mathbb{R}^d$, the $\delta$-dilation of a cube is known: $|\delta Q_j|=|\delta_1 \cdots \delta_d| \cdot |Q_j|$. So, $$\cdots = \sum_{j=1}^\infty |\delta_1 \cdots \delta_d| \cdot | Q_j| = |\delta_1 \cdots \delta_d| \sum_{j=1}^\infty |Q_j| \leq \delta_1 \cdots \delta_d \cdot \epsilon $$ Thus, $\delta E \in{\mathcal{M}}$. Also, $$m(\delta E) = m_*(\delta E) = \inf \sum_{j=1}^\infty |\delta Q_j| = \delta_1 \cdots \delta_d \cdot \inf \sum_{j=1}^\infty |Q_j|=\delta_1 \cdots \delta_d \cdot m_*(E) = \delta_1 \cdots \delta_d \cdot m(E)$$
Wednesday, May 28, 2014
Fatou's Lemma, the Monotone Convergence Theorem and the Dominated Convergence Theorem
Consider integration on $\mathbb{R}^d$ as a mapping from the vector space of measurable functions on $\mathbb{R}^d$ to $[-\infty,\infty]$. If we restrict our attention to the collection of positive measurable functions, then a natural property to examine of this mapping is continuity.
If $ f_n \longrightarrow f$ a.e. does this imply that $\int f_n \longrightarrow \int f$ ?
e.g. $f_n(x) = n \cdot \chi_{[0,\frac{1}{n}]}(x)$
Then, $\{f_n\}$ is a sequence of measurable functions, $f_n \longrightarrow 0$ a.e. and so the zero function is the pointwise limit of $\{f_n\}$. So, $\int f = \int 0 = 0$. However, $\int f_n = 1 $ $\forall n\in{\mathbb{N}}$, so we get that $\int f < \lim_{n \rightarrow \infty} \int f_n$.
In general, the sequence of extended real numbers $\{f_n\}$ may not have a limit. Any sequence of extended real numbers does always have a $\limsup$ and a $\liminf$. So, in the above example, we have, in particular that $\int f < \liminf_{n \rightarrow \infty} \int f_n$ $\textit{theorem}$: Fatou's Lemma
Given $\{f_n\}$ a sequence of positive measurable functions and $f_n \longrightarrow f$ a.e. then $$\int f \leq \liminf_{n \rightarrow \infty} \int f_n$$ $\textit{proof}$:
The sequential properties of measurable functions imply that $f$ is a positive, measurable function. By definition, the Lebesgue integral of a positive measurable function is $$\int f \equiv \sup_g \left\lbrace\int g \right\rbrace$$ where the supremum is taken over all measurable $g$, $0 \leq g \leq f$, $g$ bounded and supported on a set of finite measure. Thus, to prove Fatou's Lemma, it suffices to show that for such $g$, $$\int g \leq \liminf_{n \rightarrow \infty} \int f_n$$ Let $g$ be a function satisfying the conditions of being bounded, measurable, supported on a set of finite measure and $0 \leq g \leq f$. Define $$g_n(x) = \min(g(x),f_n(x))$$ Then, $g_n$ are measurable, $g_n \longrightarrow g$ a.e. and the monotinicity of integration implies $\int g_n \leq \int f_n$. Applying the bounded convergence theorem gives $\int g_n \longrightarrow \int g$. In particular, $$\liminf_{n \rightarrow \infty} \int g_n = \int g \leq \liminf_{n \rightarrow \infty} \int f_n$$ Then, taking the sup over all $g$ that satisfy the above conditions, we get $$\sup_g \left\lbrace \int g \right\rbrace = \int f \leq \liminf_{n \rightarrow \infty} \int f_n$$ $\blacksquare$ Note that since the above integrals may not be finite, the above inequality holds in the extended sense. That is, we may get statements of the form $\int f \leq \infty$ or $\infty \leq \infty$ This proof of Fatou's lemma uses the bounded convergence theorem; which uses Egorov's theorem. With Fatou's Lemma, one can prove the monotone convergence theorem and the dominated convergence theorem which are both used in proving that $L^1(X,\mu)$ is complete in its metric (Riesz-Fischer). Alternatively, the monotone convergence theorem may be proven independently of the above results. If we assume that the monotone convergence theorem has been proven, we may obtain an alternate version of Fatou's lemma.
\textit{Alternate Fatou's Lemma}: Assume $\{f_n\}$ is a sequence of positive measurable functions. Then, $$\int \liminf_{n \rightarrow \infty} f_n \leq \liminf_{n \rightarrow \infty} \int f_n $$ Noe that $\{f_n\}$ need not necessarily have a pointwise limit, yet we are still able to obtain the above estimate since the limit inferior of a sequence of real numbers always exists. Of course, the above inequality also holds in the extended sense. $textit{proof}$ :
Define the following sequence of functions $$g_k(x) \equiv \inf_{i \geq k} f_i(x)$$ By this definition, $g_k$ are measurable, $g_k \leq f_k$ and $g_k \leq g_{k+1}$ since inf's increase as the index variable $i$ increases. $\{g_k\}$ actually has a pointwise limit since $$\lim_{k \rightarrow \infty} g_k(x) = \lim_{k \rightarrow \infty} \inf_{i \geq k} f_i(x)$$ which is exactly the definition of the limit inferior, $\liminf f_i$, which always exists and is measurable (Stein p. 29). So by construction, $0 \leq g_k \nearrow \liminf f_i$. Applying the monotone convergence theorem, we get $$\lim_{k \rightarrow \infty} \int g_k = \int \liminf f_i$$ In particular, $\lim_{k \rightarrow \infty} \int g_k = \liminf \int g_k \leq \liminf \int f_k$. Therefore, $$\int \liminf f_i \leq \liminf \int f_i$$ $\blacksquare$
\textit{Corollary 1}: Given $\{f_n\}$ a sequence of positive measurable functions with $0 \leq f_n \leq f$ and $f_n \longrightarrow f$ a.e. then $$\int f_n \longrightarrow \int f$$ This corollary establishes some conditions in which the operations of integration and (pointwise) limit may be swapped.
$\textit{proof}$ :
Note that the above conditions are slightly stronger than those of Fatou's lemma. So, applying Fatou's lemma gives us that $\int f \leq \liminf_{n \rightarrow \infty} \int f_n$. On the other hand, with $f_n \leq f$, the monotinicity of integration gives us that $\int f_n \leq \int f$. A priori, $\lim_{n \rightarrow \infty} \int f_n$ may not exist, but $\limsup_{n \rightarrow \infty} \int f_n$ does and $\limsup_{n \rightarrow \infty} \int f_n \leq \int f$. We now use the fact that for any sequence $\{a_n\}$ of real numbers, $\liminf_{n \rightarrow \infty} \{a_n\} \leq \limsup_{n \rightarrow \infty} \{a_n\}$. Since $\left\lbrace \int f_n \right\rbrace$ is really just a sequence of real numbers, we get $$\int f \leq \liminf_{n \rightarrow \infty} \int f_n \leq \limsup_{n \rightarrow \infty} \int f_n \leq \int f$$ So, $\liminf_{n \rightarrow \infty} \int f_n = \limsup_{n \rightarrow \infty} \int f_n \Longrightarrow \lim_{n \rightarrow \infty} \int f_n$ exists and equals $\int f$. $\blacksquare$ \textit{Corollary 2}: If $\{f_n\}$ is a sequence of measurable functions, $f_n \longrightarrow f$ a.e. and $\exists h \in{L^1}$ s.t. $-h \leq f_n$ $\forall n \in{\mathbb{N}}$ then $\int f \leq \liminf \int f_n$. If additionally, $f_n \leq f$ then again, $\int f_n \longrightarrow \int f$ $\textit{proof}$ :
Since $-h \leq f_n$, then adding $h$ to both sides of the inequality yields $0 \leq f_n+h$. Since $h\in{L^1}$, $h$ is, in particular, measurable. So, $\{f_n+h\}$ is a sequence of positive measurable functions converging a.e. to $f+h$. Applying Fatou's lemma gives $\int f+h \leq \liminf \int(f_n + h) = \liminf \int f_n + \int h$. Since $h\in{L^1}$, $\int h \leq \int |h| < \infty$, so subtracting $\int h$ from both sides gives $\int f \leq \liminf \int f_n$.
With the additional assumption that $f_n \leq f$, then we get that $-h \leq f_n \leq f$ so $0 \leq f_n + h \leq f+h$. Applying the previous corollary to the sequence $\{f_n+h\}$ gives us the interchange of limit and integration result. $\blacksquare$ $\textit{theorem}$ :
$\textit{Monotone Convergence Theorem}$
Given $\{f_n\}$ a sequence of positive measurable functions with $0 \leq f_n \nearrow f$ a.e. then $$\int f_n \longrightarrow \int f$$ The above conditions are stronger still than those of corollary 1, so there is nothing to prove.
The next corollary to the monotone convergence theorem is a result for series.
\textit{Corollary}: Assume that $\{a_j\}$ is a positive sequence of measurable functions. Consider the series $\sum_{j=1}^\infty a_j$. Then, $$\int \sum_{j=1}^\infty a_j = \sum_{j=1}^\infty \int a_j $$ In general, the above equality holds in the extended sense and the series may not converge. However, if the above equality holds in the finite (non-extended) sense, then the integrand function $\sum_{j=1}^\infty a_j$ must be finite a.e. That is, the series $\sum_{j=1}^\infty a_j$ converges a.e.
Another interpretation is that this corollary establishes conditions in which the linearity of the integral may be generalized to countable sums. $\textit{proof}$ :
We wish to construct a sequence of functions satisfying the conditions of the monotone convergence theorem. Define $$f_N = \sum_{j=1}^Na_j$$ Then, since $a_j \geq 0$, adding more terms to the above series only increases the value. So, $ 0 \leq f_1 \leq f_2 \leq \cdots \leq f_N \leq f_{N+1} \leq \cdots \leq f = \sum_{j=1}^\infty a_j$. The monotone convergence theorem implies that $$\lim_{N \rightarrow \infty} \int f_N = \int f$$ i.e. $$\lim_{N \rightarrow \infty} \int \sum_{j=1}^N a_j = \int \sum_{j=1}^\infty a_j$$ Since the Lebesgue integral is linear for a finite number of added terms, $$ \cdots = \lim_{N \rightarrow \infty} \sum_{j=1}^N a_j \int = \sum_{j=1}^\infty \int a_j$$ $\blacksquare$ This result will be a useful tool in the proof of Riesz-Fischer.
So, we have seen that pointwise convergence is not a sufficient condition to allow us to interchange the operations of limit and integration. What kinds of conditions do we need? Consider the following examples:
1. $f_n = \frac{1}{n}$ $\forall n \in{\mathbb{N}}$. Then, $f_n \longrightarrow 0$ (actually uniformly on $\mathbb{R}$) however, $\int_{\mathbb{R}} f_n = + \infty$ $\forall n \in{\mathbb{N}}$ whereas $\int_{\mathbb{R}} 0 = 0$.
2. $f_n = \frac{1}{n} \chi_{[\frac{-n}{2},\frac{n}{2}]}$ also converges uniformly to 0 on all of $\mathbb{R}$. $\int_{\mathbb{R}} f_n = 1$ $\forall n \in{\mathbb{N}}$
3. $f_n : [0,1] \longrightarrow \mathbb{R}$ converges uniformly to zero, is it true that $\lim_{n \rightarrow \infty} \int_{\mathbb{R}}f_n = 0?$ Yes, by the monotinicity of the integral, $\int_{[0,1]} f_n \leq \sup_{0 \leq x \leq 1}|f_n| \longrightarrow 0$ as $n \rightarrow \infty$.
The third example also illustrates an important fact: (Everywhere) uniform convergence forces convergence in the sup norm, which can in turn force bounds on integrals. The key difference in the third example is that each of the $f_n$ can be "dominated" by a single function which is $L^1$.
The next theorem is one of the most useful convergence results. $\textit{theorem}$ : Lebesgue Dominated Convergence Theorem
If $\{f_n\}$ is a sequence of measurable functions s.t. $|f_n| \leq g \in{(L^1,\mathbb{R}^d)}$ and $f_n \longrightarrow f$ a.e. then $$\int |f_n-f| \rightarrow 0$$ If we have the notion of the normed space $(L^1, ||\cdot||_1)$ then the conclusion of the DCT can be regarded as $||f_n-f||_1 \longrightarrow 0$. So in another sense, the DCT can be regarded as a theorem that establishes conditions in which pointwise convergence becomes convergence in the $L^1$ norm. For general measurable functions, neither this implication nor its converse are true. (Stein 2.8)
Example problem: Suppose $f_n: \mathbb{R}^d \longrightarrow [0,\infty]$ are measurable $\forall n\in{\mathbb{N}}$, $f_1 \geq f_2 \geq f_3 \geq \cdots \geq 0 $, $f_n \longrightarrow f$ a.e. and $f_1 \in{L^1}$. Show that $\lim_{n \rightarrow \infty} \int f_n = \int f$.
The dominated convergence theorem implies that $$\int |f_n-f| \longrightarrow 0$$ Using the triangle inequality and linearity of the integral, $$0 \leq \left| \int f_n - \int f \right|=\left| \int (f_n-f)\right| \leq \int |f_n-f| \longrightarrow 0$$ forces $$\int f_n \longrightarrow \int f$$ $\textit{proof}$ :
To show that $ \lim_{n \rightarrow \infty} \int |f_n-f|$ actually exists and equals zero, we will show that $\liminf \int |f_n-f| = \limsup \int |f_n-f| = 0$. This suggests that we should use Fatou's lemma in some way. Since $|f_n| \leq g$, then certainly $2g + |f_n-f| \geq 0$ and $2g-|f_n-f| \geq 0$ . Since $g\in{L^1}$, $g$ is, in particular, measurable. So, $2g \pm |f_n-f|$ being an algebraic combination of measurable functions is measurable. Also, since $f_n \longrightarrow f$ a.e. then $|f_n-f| \longrightarrow 0$ a.e. , so $2g \pm |f_n-f| \longrightarrow 2g$ a.e. Applying Fatou's lemma, $$0 \leq \int 2g \leq \liminf \int (2g + |f_n-f|) = \liminf \left( \int 2g + \int |f_n-f| \right) $$ where we are using the linearity of the integral. Recall some important properties of $\liminf $ and $\limsup$: If $\{a_n \}$ is a sequence of real numbers and $c$ is a constant, then $\liminf(c+a_n) = c+ \liminf a_n$. Since $\left\lbrace \int |f_n-f| \right\rbrace $ is really just a sequence of real numbers, then $$ \liminf \left( \int 2g + \int |f_n-f| \right) = \int 2g + \liminf \int |f_n-f|$$ Since $g\in{L^1}$, $\int 2g < \infty$ and so we are permitted to subtract $\int 2g$ from both sides of the above inequality to get $$0 \leq \liminf \int |f_n-f| $$ On the other hand, applying Fatou's lemma to the sequence $2g-|f_n-f|$, we get $$0 \leq \int 2g \leq \liminf \int (2g-|f_n-f|) = \int 2g + \liminf \left( -\int |f_n-f| \right) $$ $$0 \leq \liminf \left( -\int |f_n-f| \right)$$ We will now use two more properties of $\liminf$ and $\limsup$:
1. $\liminf a_n = - \limsup (-a_n)$
2. $\liminf a_n \leq \limsup a_n $
So, $$0 \leq \liminf \left( -\int |f_n-f| \right) = - \limsup \int |f_n-f|$$ $$0 \geq \limsup \int |f_n-f|$$ The final property of $\liminf$ and $\limsup$ we use is that $\liminf a_n \leq \limsup a_n$. Combining this property with (1) and (2), we obtain $$0 \leq \liminf \int |f_n-f| \leq \limsup \int |f_n-f| \leq 0$$ Which immediately implies that $\liminf \int |f_n-f| = \limsup \int |f_n-f|$ and so $\lim_{n \rightarrow \infty} \int |f_n-f|$ exists and equals 0.
If $ f_n \longrightarrow f$ a.e. does this imply that $\int f_n \longrightarrow \int f$ ?
e.g. $f_n(x) = n \cdot \chi_{[0,\frac{1}{n}]}(x)$
Then, $\{f_n\}$ is a sequence of measurable functions, $f_n \longrightarrow 0$ a.e. and so the zero function is the pointwise limit of $\{f_n\}$. So, $\int f = \int 0 = 0$. However, $\int f_n = 1 $ $\forall n\in{\mathbb{N}}$, so we get that $\int f < \lim_{n \rightarrow \infty} \int f_n$.
In general, the sequence of extended real numbers $\{f_n\}$ may not have a limit. Any sequence of extended real numbers does always have a $\limsup$ and a $\liminf$. So, in the above example, we have, in particular that $\int f < \liminf_{n \rightarrow \infty} \int f_n$ $\textit{theorem}$: Fatou's Lemma
Given $\{f_n\}$ a sequence of positive measurable functions and $f_n \longrightarrow f$ a.e. then $$\int f \leq \liminf_{n \rightarrow \infty} \int f_n$$ $\textit{proof}$:
The sequential properties of measurable functions imply that $f$ is a positive, measurable function. By definition, the Lebesgue integral of a positive measurable function is $$\int f \equiv \sup_g \left\lbrace\int g \right\rbrace$$ where the supremum is taken over all measurable $g$, $0 \leq g \leq f$, $g$ bounded and supported on a set of finite measure. Thus, to prove Fatou's Lemma, it suffices to show that for such $g$, $$\int g \leq \liminf_{n \rightarrow \infty} \int f_n$$ Let $g$ be a function satisfying the conditions of being bounded, measurable, supported on a set of finite measure and $0 \leq g \leq f$. Define $$g_n(x) = \min(g(x),f_n(x))$$ Then, $g_n$ are measurable, $g_n \longrightarrow g$ a.e. and the monotinicity of integration implies $\int g_n \leq \int f_n$. Applying the bounded convergence theorem gives $\int g_n \longrightarrow \int g$. In particular, $$\liminf_{n \rightarrow \infty} \int g_n = \int g \leq \liminf_{n \rightarrow \infty} \int f_n$$ Then, taking the sup over all $g$ that satisfy the above conditions, we get $$\sup_g \left\lbrace \int g \right\rbrace = \int f \leq \liminf_{n \rightarrow \infty} \int f_n$$ $\blacksquare$ Note that since the above integrals may not be finite, the above inequality holds in the extended sense. That is, we may get statements of the form $\int f \leq \infty$ or $\infty \leq \infty$ This proof of Fatou's lemma uses the bounded convergence theorem; which uses Egorov's theorem. With Fatou's Lemma, one can prove the monotone convergence theorem and the dominated convergence theorem which are both used in proving that $L^1(X,\mu)$ is complete in its metric (Riesz-Fischer). Alternatively, the monotone convergence theorem may be proven independently of the above results. If we assume that the monotone convergence theorem has been proven, we may obtain an alternate version of Fatou's lemma.
\textit{Alternate Fatou's Lemma}: Assume $\{f_n\}$ is a sequence of positive measurable functions. Then, $$\int \liminf_{n \rightarrow \infty} f_n \leq \liminf_{n \rightarrow \infty} \int f_n $$ Noe that $\{f_n\}$ need not necessarily have a pointwise limit, yet we are still able to obtain the above estimate since the limit inferior of a sequence of real numbers always exists. Of course, the above inequality also holds in the extended sense. $textit{proof}$ :
Define the following sequence of functions $$g_k(x) \equiv \inf_{i \geq k} f_i(x)$$ By this definition, $g_k$ are measurable, $g_k \leq f_k$ and $g_k \leq g_{k+1}$ since inf's increase as the index variable $i$ increases. $\{g_k\}$ actually has a pointwise limit since $$\lim_{k \rightarrow \infty} g_k(x) = \lim_{k \rightarrow \infty} \inf_{i \geq k} f_i(x)$$ which is exactly the definition of the limit inferior, $\liminf f_i$, which always exists and is measurable (Stein p. 29). So by construction, $0 \leq g_k \nearrow \liminf f_i$. Applying the monotone convergence theorem, we get $$\lim_{k \rightarrow \infty} \int g_k = \int \liminf f_i$$ In particular, $\lim_{k \rightarrow \infty} \int g_k = \liminf \int g_k \leq \liminf \int f_k$. Therefore, $$\int \liminf f_i \leq \liminf \int f_i$$ $\blacksquare$
\textit{Corollary 1}: Given $\{f_n\}$ a sequence of positive measurable functions with $0 \leq f_n \leq f$ and $f_n \longrightarrow f$ a.e. then $$\int f_n \longrightarrow \int f$$ This corollary establishes some conditions in which the operations of integration and (pointwise) limit may be swapped.
$\textit{proof}$ :
Note that the above conditions are slightly stronger than those of Fatou's lemma. So, applying Fatou's lemma gives us that $\int f \leq \liminf_{n \rightarrow \infty} \int f_n$. On the other hand, with $f_n \leq f$, the monotinicity of integration gives us that $\int f_n \leq \int f$. A priori, $\lim_{n \rightarrow \infty} \int f_n$ may not exist, but $\limsup_{n \rightarrow \infty} \int f_n$ does and $\limsup_{n \rightarrow \infty} \int f_n \leq \int f$. We now use the fact that for any sequence $\{a_n\}$ of real numbers, $\liminf_{n \rightarrow \infty} \{a_n\} \leq \limsup_{n \rightarrow \infty} \{a_n\}$. Since $\left\lbrace \int f_n \right\rbrace$ is really just a sequence of real numbers, we get $$\int f \leq \liminf_{n \rightarrow \infty} \int f_n \leq \limsup_{n \rightarrow \infty} \int f_n \leq \int f$$ So, $\liminf_{n \rightarrow \infty} \int f_n = \limsup_{n \rightarrow \infty} \int f_n \Longrightarrow \lim_{n \rightarrow \infty} \int f_n$ exists and equals $\int f$. $\blacksquare$ \textit{Corollary 2}: If $\{f_n\}$ is a sequence of measurable functions, $f_n \longrightarrow f$ a.e. and $\exists h \in{L^1}$ s.t. $-h \leq f_n$ $\forall n \in{\mathbb{N}}$ then $\int f \leq \liminf \int f_n$. If additionally, $f_n \leq f$ then again, $\int f_n \longrightarrow \int f$ $\textit{proof}$ :
Since $-h \leq f_n$, then adding $h$ to both sides of the inequality yields $0 \leq f_n+h$. Since $h\in{L^1}$, $h$ is, in particular, measurable. So, $\{f_n+h\}$ is a sequence of positive measurable functions converging a.e. to $f+h$. Applying Fatou's lemma gives $\int f+h \leq \liminf \int(f_n + h) = \liminf \int f_n + \int h$. Since $h\in{L^1}$, $\int h \leq \int |h| < \infty$, so subtracting $\int h$ from both sides gives $\int f \leq \liminf \int f_n$.
With the additional assumption that $f_n \leq f$, then we get that $-h \leq f_n \leq f$ so $0 \leq f_n + h \leq f+h$. Applying the previous corollary to the sequence $\{f_n+h\}$ gives us the interchange of limit and integration result. $\blacksquare$ $\textit{theorem}$ :
$\textit{Monotone Convergence Theorem}$
Given $\{f_n\}$ a sequence of positive measurable functions with $0 \leq f_n \nearrow f$ a.e. then $$\int f_n \longrightarrow \int f$$ The above conditions are stronger still than those of corollary 1, so there is nothing to prove.
The next corollary to the monotone convergence theorem is a result for series.
\textit{Corollary}: Assume that $\{a_j\}$ is a positive sequence of measurable functions. Consider the series $\sum_{j=1}^\infty a_j$. Then, $$\int \sum_{j=1}^\infty a_j = \sum_{j=1}^\infty \int a_j $$ In general, the above equality holds in the extended sense and the series may not converge. However, if the above equality holds in the finite (non-extended) sense, then the integrand function $\sum_{j=1}^\infty a_j$ must be finite a.e. That is, the series $\sum_{j=1}^\infty a_j$ converges a.e.
Another interpretation is that this corollary establishes conditions in which the linearity of the integral may be generalized to countable sums. $\textit{proof}$ :
We wish to construct a sequence of functions satisfying the conditions of the monotone convergence theorem. Define $$f_N = \sum_{j=1}^Na_j$$ Then, since $a_j \geq 0$, adding more terms to the above series only increases the value. So, $ 0 \leq f_1 \leq f_2 \leq \cdots \leq f_N \leq f_{N+1} \leq \cdots \leq f = \sum_{j=1}^\infty a_j$. The monotone convergence theorem implies that $$\lim_{N \rightarrow \infty} \int f_N = \int f$$ i.e. $$\lim_{N \rightarrow \infty} \int \sum_{j=1}^N a_j = \int \sum_{j=1}^\infty a_j$$ Since the Lebesgue integral is linear for a finite number of added terms, $$ \cdots = \lim_{N \rightarrow \infty} \sum_{j=1}^N a_j \int = \sum_{j=1}^\infty \int a_j$$ $\blacksquare$ This result will be a useful tool in the proof of Riesz-Fischer.
So, we have seen that pointwise convergence is not a sufficient condition to allow us to interchange the operations of limit and integration. What kinds of conditions do we need? Consider the following examples:
1. $f_n = \frac{1}{n}$ $\forall n \in{\mathbb{N}}$. Then, $f_n \longrightarrow 0$ (actually uniformly on $\mathbb{R}$) however, $\int_{\mathbb{R}} f_n = + \infty$ $\forall n \in{\mathbb{N}}$ whereas $\int_{\mathbb{R}} 0 = 0$.
2. $f_n = \frac{1}{n} \chi_{[\frac{-n}{2},\frac{n}{2}]}$ also converges uniformly to 0 on all of $\mathbb{R}$. $\int_{\mathbb{R}} f_n = 1$ $\forall n \in{\mathbb{N}}$
3. $f_n : [0,1] \longrightarrow \mathbb{R}$ converges uniformly to zero, is it true that $\lim_{n \rightarrow \infty} \int_{\mathbb{R}}f_n = 0?$ Yes, by the monotinicity of the integral, $\int_{[0,1]} f_n \leq \sup_{0 \leq x \leq 1}|f_n| \longrightarrow 0$ as $n \rightarrow \infty$.
The third example also illustrates an important fact: (Everywhere) uniform convergence forces convergence in the sup norm, which can in turn force bounds on integrals. The key difference in the third example is that each of the $f_n$ can be "dominated" by a single function which is $L^1$.
The next theorem is one of the most useful convergence results. $\textit{theorem}$ : Lebesgue Dominated Convergence Theorem
If $\{f_n\}$ is a sequence of measurable functions s.t. $|f_n| \leq g \in{(L^1,\mathbb{R}^d)}$ and $f_n \longrightarrow f$ a.e. then $$\int |f_n-f| \rightarrow 0$$ If we have the notion of the normed space $(L^1, ||\cdot||_1)$ then the conclusion of the DCT can be regarded as $||f_n-f||_1 \longrightarrow 0$. So in another sense, the DCT can be regarded as a theorem that establishes conditions in which pointwise convergence becomes convergence in the $L^1$ norm. For general measurable functions, neither this implication nor its converse are true. (Stein 2.8)
Example problem: Suppose $f_n: \mathbb{R}^d \longrightarrow [0,\infty]$ are measurable $\forall n\in{\mathbb{N}}$, $f_1 \geq f_2 \geq f_3 \geq \cdots \geq 0 $, $f_n \longrightarrow f$ a.e. and $f_1 \in{L^1}$. Show that $\lim_{n \rightarrow \infty} \int f_n = \int f$.
The dominated convergence theorem implies that $$\int |f_n-f| \longrightarrow 0$$ Using the triangle inequality and linearity of the integral, $$0 \leq \left| \int f_n - \int f \right|=\left| \int (f_n-f)\right| \leq \int |f_n-f| \longrightarrow 0$$ forces $$\int f_n \longrightarrow \int f$$ $\textit{proof}$ :
To show that $ \lim_{n \rightarrow \infty} \int |f_n-f|$ actually exists and equals zero, we will show that $\liminf \int |f_n-f| = \limsup \int |f_n-f| = 0$. This suggests that we should use Fatou's lemma in some way. Since $|f_n| \leq g$, then certainly $2g + |f_n-f| \geq 0$ and $2g-|f_n-f| \geq 0$ . Since $g\in{L^1}$, $g$ is, in particular, measurable. So, $2g \pm |f_n-f|$ being an algebraic combination of measurable functions is measurable. Also, since $f_n \longrightarrow f$ a.e. then $|f_n-f| \longrightarrow 0$ a.e. , so $2g \pm |f_n-f| \longrightarrow 2g$ a.e. Applying Fatou's lemma, $$0 \leq \int 2g \leq \liminf \int (2g + |f_n-f|) = \liminf \left( \int 2g + \int |f_n-f| \right) $$ where we are using the linearity of the integral. Recall some important properties of $\liminf $ and $\limsup$: If $\{a_n \}$ is a sequence of real numbers and $c$ is a constant, then $\liminf(c+a_n) = c+ \liminf a_n$. Since $\left\lbrace \int |f_n-f| \right\rbrace $ is really just a sequence of real numbers, then $$ \liminf \left( \int 2g + \int |f_n-f| \right) = \int 2g + \liminf \int |f_n-f|$$ Since $g\in{L^1}$, $\int 2g < \infty$ and so we are permitted to subtract $\int 2g$ from both sides of the above inequality to get $$0 \leq \liminf \int |f_n-f| $$ On the other hand, applying Fatou's lemma to the sequence $2g-|f_n-f|$, we get $$0 \leq \int 2g \leq \liminf \int (2g-|f_n-f|) = \int 2g + \liminf \left( -\int |f_n-f| \right) $$ $$0 \leq \liminf \left( -\int |f_n-f| \right)$$ We will now use two more properties of $\liminf$ and $\limsup$:
1. $\liminf a_n = - \limsup (-a_n)$
2. $\liminf a_n \leq \limsup a_n $
So, $$0 \leq \liminf \left( -\int |f_n-f| \right) = - \limsup \int |f_n-f|$$ $$0 \geq \limsup \int |f_n-f|$$ The final property of $\liminf$ and $\limsup$ we use is that $\liminf a_n \leq \limsup a_n$. Combining this property with (1) and (2), we obtain $$0 \leq \liminf \int |f_n-f| \leq \limsup \int |f_n-f| \leq 0$$ Which immediately implies that $\liminf \int |f_n-f| = \limsup \int |f_n-f|$ and so $\lim_{n \rightarrow \infty} \int |f_n-f|$ exists and equals 0.
Tuesday, May 27, 2014
Continuity of Measure (Stein Cor. 3.3)
Let $\{E_j\}$ be a sequence of Lebesgue measurable sets of $\mathbb{R}^d$. Then
(i) If $E_j \nearrow E = \cup_{J=1}^\infty $ then $m(E_j) \longrightarrow m(E)$
(ii) If $E_j \searrow E = \cup_{J=1}^\infty $ and $m(E_k) < \infty$ for some $k\in{\mathbb{N}}$ then $m(E_j) \longrightarrow m(E)$.JPG)
.JPG)
(i) If $E_j \nearrow E = \cup_{J=1}^\infty $ then $m(E_j) \longrightarrow m(E)$
(ii) If $E_j \searrow E = \cup_{J=1}^\infty $ and $m(E_k) < \infty$ for some $k\in{\mathbb{N}}$ then $m(E_j) \longrightarrow m(E)$
.JPG)
.JPG)
Saturday, May 17, 2014
Norms: a short overview
Recall: A $\textit{norm}$ (in the sense of analysis) is a map from a vector space to the non-negative real numbers
$|| \cdot ||: V \longrightarrow [0,\infty)$ with the following 3 properties :
(i) $||x|| \geq 0 \quad \forall{x \in{V}}$ and $||x||=0 \Longleftrightarrow x=0$ (positive-definite)
(ii) $||\alpha x||=|\alpha| \cdot ||x||$ $\forall \alpha\in{F}$, $x\in{V}$ (homogeneity)
(iii) $||x+y|| \leq ||x||+||y||$ (triangle inequality)
Any vector space equipped with a mapping satisfying (i)-(iii) above is called a $\textit{normed space}$
Examples:
1. The real numbers as a vector space together with the absolute value function $(\mathbb{R},|\cdot |)$ form a classic example of a normed space. Many other mappings that are defined by taking the absolute value of a real number inherit (i)-(iii) above and are thus a norm.
2. Consider the vector space of real-valued continuous functions defined on a closed interval $[a,b]$ (denoted $C[a,b]$). The norm $$||f||_p = \left(\int_a^b|f(x)|^p dx \right)^{\frac{1}{p}} \quad p\in{[1,\infty)}$$ When $p=1$, the above defines both a norm and a linear functional by the linearity and triangle inequality of the integral. The triangle inequality for the cases $ 1 < p < \infty $ is called Minkowski's inequality. For the $ p= \infty $ case, $$||f||_\infty \equiv \inf_{a \leq x \leq b}\{M \geq 0 : |f(x)| \leq M \quad a.e. \}$$ Since real-valued continuous functions on a closed bounded interval attain their sup and inf, the above norm is equivalent to $$||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)| $$ When $C[a,b]$ is equipped with the norm $||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)|$ this normed space is $\textit{complete}$: Sequences which are Cauchy with respect to the max norm also converge to an element in $C[a,b]$.
3. The mapping $||f||_1 \equiv \int_X|f(x)|d\mu(x)$ forms a $\textit{pseudonorm}$ on the vector space of integrable functions $L^1(X, \mu)$. Regarding (i), we may have $||f||=0$ for some $f$ that is not identically zero as long as $f = 0$ $\mu$-almost everywhere. e.g. The Dirichlet function $\chi_{\mathbb{Q}}=0$ $m-a.e.$ where $m$ is the usual Lebesgue measure on $\mathbb{R}$. To make a norm, we define the equivalence relation $f$~$g$ if $\mu(\{x|f(x) \neq g(x)\})=0$. Then, the set of equivalence classes of $L^1$ functions together with $|| \cdot ||$ forms a normed space.
Regarding (ii) above, the homogeneity requirement of norms has important implications:
$\textit{The powers of the norms must agree} $: Consider the following inequality
$||f||_X^2 \leq c||f||_Y$ If this inequality is true for all $f$, then in particular it is true if we replace $f$ with $\alpha f$.
$||\alpha f||_X^2 \leq c ||\alpha f||_Y$
$(|\alpha| ||f||_X)^2 \leq c |\alpha| ||f||_Y$
$|\alpha|^2 ||f||^2 \leq c |\alpha| ||f||_Y$
$|\alpha| ||f||_X^2 \leq c||f||_Y$
Letting $\alpha \longrightarrow 0$, we get $0 \leq c||f||_Y$...which is trivial. Also,
$||f||_X^2 \leq \frac{c}{|\alpha|} ||f||_Y$. Letting $\alpha \longrightarrow + \infty$, we get $||f||_X^2 = 0 \Longleftrightarrow f = 0$...nonsense!
Additionally, $\textit{the powers of the internal constants must agree}$. Consider the inequality
$||\alpha f|| \leq c||f||$. If true for all $f$, then again replace $f$ with $\alpha f$
$||\alpha \alpha f|| \leq c|| \alpha f|| $
$ | \alpha|^2 ||f|| \leq c |\alpha| ||f||$
$||f|| \leq \frac{c}{|\alpha |} ||f||$. Letting $| \alpha | \longrightarrow + \infty $ , we get
$||f|| = 0$...nonsense!
Regarding (iii) norms also satisfy the reverse triangle inequality: $$\left| ||x||-||y|| \right| \leq ||x-y|| \quad \forall x,y \in{(V,|| \cdot ||)} $$ Let $x,y$ be in a normed space and let $||x|| > ||y||$ for if they are equal, then the reverse triangle inequality is trivial.
$||x+y|| \leq ||x|| + ||y||$
$ ||x+y|| - ||y|| \leq ||x||$. $\leftarrow$ since this is true forall $x,y$, it is also true if we replace $x$ with $x-y$.
$||(x-y)+y|| - ||y|| \leq ||x-y||$
$||x||-||y|| \leq ||x-y||$. With both sides being non-negative, we may put absolute values on the left hand side
$\left| ||x||-||y|| \right| \leq ||x-y||$
Note that if we started with the above line, we could have reversed each of the above steps to obtain the triangle inequality. So, these two inequalities are equivalent in any normed space.
$|| \cdot ||: V \longrightarrow [0,\infty)$ with the following 3 properties :
(i) $||x|| \geq 0 \quad \forall{x \in{V}}$ and $||x||=0 \Longleftrightarrow x=0$ (positive-definite)
(ii) $||\alpha x||=|\alpha| \cdot ||x||$ $\forall \alpha\in{F}$, $x\in{V}$ (homogeneity)
(iii) $||x+y|| \leq ||x||+||y||$ (triangle inequality)
Any vector space equipped with a mapping satisfying (i)-(iii) above is called a $\textit{normed space}$
Examples:
1. The real numbers as a vector space together with the absolute value function $(\mathbb{R},|\cdot |)$ form a classic example of a normed space. Many other mappings that are defined by taking the absolute value of a real number inherit (i)-(iii) above and are thus a norm.
2. Consider the vector space of real-valued continuous functions defined on a closed interval $[a,b]$ (denoted $C[a,b]$). The norm $$||f||_p = \left(\int_a^b|f(x)|^p dx \right)^{\frac{1}{p}} \quad p\in{[1,\infty)}$$ When $p=1$, the above defines both a norm and a linear functional by the linearity and triangle inequality of the integral. The triangle inequality for the cases $ 1 < p < \infty $ is called Minkowski's inequality. For the $ p= \infty $ case, $$||f||_\infty \equiv \inf_{a \leq x \leq b}\{M \geq 0 : |f(x)| \leq M \quad a.e. \}$$ Since real-valued continuous functions on a closed bounded interval attain their sup and inf, the above norm is equivalent to $$||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)| $$ When $C[a,b]$ is equipped with the norm $||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)|$ this normed space is $\textit{complete}$: Sequences which are Cauchy with respect to the max norm also converge to an element in $C[a,b]$.
3. The mapping $||f||_1 \equiv \int_X|f(x)|d\mu(x)$ forms a $\textit{pseudonorm}$ on the vector space of integrable functions $L^1(X, \mu)$. Regarding (i), we may have $||f||=0$ for some $f$ that is not identically zero as long as $f = 0$ $\mu$-almost everywhere. e.g. The Dirichlet function $\chi_{\mathbb{Q}}=0$ $m-a.e.$ where $m$ is the usual Lebesgue measure on $\mathbb{R}$. To make a norm, we define the equivalence relation $f$~$g$ if $\mu(\{x|f(x) \neq g(x)\})=0$. Then, the set of equivalence classes of $L^1$ functions together with $|| \cdot ||$ forms a normed space.
Regarding (ii) above, the homogeneity requirement of norms has important implications:
$\textit{The powers of the norms must agree} $: Consider the following inequality
$||f||_X^2 \leq c||f||_Y$ If this inequality is true for all $f$, then in particular it is true if we replace $f$ with $\alpha f$.
$||\alpha f||_X^2 \leq c ||\alpha f||_Y$
$(|\alpha| ||f||_X)^2 \leq c |\alpha| ||f||_Y$
$|\alpha|^2 ||f||^2 \leq c |\alpha| ||f||_Y$
$|\alpha| ||f||_X^2 \leq c||f||_Y$
Letting $\alpha \longrightarrow 0$, we get $0 \leq c||f||_Y$...which is trivial. Also,
$||f||_X^2 \leq \frac{c}{|\alpha|} ||f||_Y$. Letting $\alpha \longrightarrow + \infty$, we get $||f||_X^2 = 0 \Longleftrightarrow f = 0$...nonsense!
Additionally, $\textit{the powers of the internal constants must agree}$. Consider the inequality
$||\alpha f|| \leq c||f||$. If true for all $f$, then again replace $f$ with $\alpha f$
$||\alpha \alpha f|| \leq c|| \alpha f|| $
$ | \alpha|^2 ||f|| \leq c |\alpha| ||f||$
$||f|| \leq \frac{c}{|\alpha |} ||f||$. Letting $| \alpha | \longrightarrow + \infty $ , we get
$||f|| = 0$...nonsense!
Regarding (iii) norms also satisfy the reverse triangle inequality: $$\left| ||x||-||y|| \right| \leq ||x-y|| \quad \forall x,y \in{(V,|| \cdot ||)} $$ Let $x,y$ be in a normed space and let $||x|| > ||y||$ for if they are equal, then the reverse triangle inequality is trivial.
$||x+y|| \leq ||x|| + ||y||$
$ ||x+y|| - ||y|| \leq ||x||$. $\leftarrow$ since this is true forall $x,y$, it is also true if we replace $x$ with $x-y$.
$||(x-y)+y|| - ||y|| \leq ||x-y||$
$||x||-||y|| \leq ||x-y||$. With both sides being non-negative, we may put absolute values on the left hand side
$\left| ||x||-||y|| \right| \leq ||x-y||$
Note that if we started with the above line, we could have reversed each of the above steps to obtain the triangle inequality. So, these two inequalities are equivalent in any normed space.
Tuesday, May 13, 2014
Application of the Arzela-Ascoli theorem
$\textbf{Application of the Arzel{\'a}-Ascoli Theorem:}$
Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on [0,1] satisfying $\int_0^1 f_n^2(t)dt \leq M$ $\forall n\in{\mathbb{N}}$ and some $M \geq 0$. Define $g_n(x) = \int_0^xf_n(t)dt$. Show that $\exists \{g_{n_k}\} \subseteq \{g_n\}$ s.t. $g_{n_k} \longrightarrow g$ uniformly.
$\textit{Recall}$ : The Arzel{\'a}-Ascoli theorem states that if $\{g_n\}$ is an equicontinuous sequence of real-valued functions defined on a compact metric space that are uniformly bounded: $|g_n(x)| \leq M$ $\forall n$ and some $M \geq 0$, then, $\exists \{g_{n_k}\}$ s.t. $g_{n_k} \longrightarrow g\in{C(X)}$ uniformly.
$([0,1],|\cdot | )$ is a compact metric space...check. So, if we can show that $\{g_n\}$ is uniformly bounded and equicontinuous, then the Arzel{\'a}-Ascoli theorem will apply and give us the desired conclusion.
We now want to show that $\{g_n\}$ is uniformly bounded: By assumption, $f_n\in{L^2[0,1]}$. Let $h=1$ on $[0,1]$. Then H{\"o}lder's inequality with p=q=2 (Cauchy-Schwarz) gives us that $$||f_n \cdot 1||_1 \leq ||f_n||_2 \cdot ||1||_2$$ $$\int_0^1|f_n \cdot 1| \leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}} \left(\int_0^1|1^2|\right)^{\frac{1}{2}} = \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}$$ Squaring both sides, $$\left(\int_0^1|f_n|\right)^2 \leq \int_0^1|f_n|^2 \leq M$$ Thus, $f\in{L^1[0,1]}$. Note that we actually showed that $L^p[a,b] \subseteq L^1[a,b] \quad \forall 1 \leq p < \infty$. To compare these inequalities with the $g_n$ terms, $$|g_n|^2=\left|\int_0^xf_n\right|^2 \leq \left(\int_0^x|f_n|\right)^2 \leq \left(\int_0^1|f_n|\right)^2 \leq M$$ $$|g_n|^2 \leq M \Longrightarrow |g_n| \leq \sqrt{M}$$ Thus, $\{g_n\}$ is a uniformly bounded sequence...check.
Now consider $0\leq x < z \leq 1$. $$|g_n(x)-g_n(z)|=\left|\int_0^xf_n - \int_0^z f_n \right|=\left|\int_x^zf_n\right|\leq \int_x^z|f_n \cdot 1|$$ Using H{\"o}lder's inequality again, $$\leq \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\left(\int_x^z|1|^2\right)^{\frac{1}{2}} = \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|}\leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|} $$ $$ \leq \sqrt[4]{M}\sqrt{|x-z|}$$ So, $|g_n(x)-g_n(z)| \leq \sqrt[4]{M}\sqrt{|x-z|}$ which gives us a Lipschitz condition on $g_n$ $\forall n$. In the $\epsilon-\delta$ criterion for equicontinuity, if we choose $\delta \leq \frac{\epsilon^2}{\sqrt{M}}$ then $|g_n(x)-g_n(z)| \leq \epsilon$ making $\{g_n\}$ uniformly equicontinuous.
Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on [0,1] satisfying $\int_0^1 f_n^2(t)dt \leq M$ $\forall n\in{\mathbb{N}}$ and some $M \geq 0$. Define $g_n(x) = \int_0^xf_n(t)dt$. Show that $\exists \{g_{n_k}\} \subseteq \{g_n\}$ s.t. $g_{n_k} \longrightarrow g$ uniformly.
$\textit{Recall}$ : The Arzel{\'a}-Ascoli theorem states that if $\{g_n\}$ is an equicontinuous sequence of real-valued functions defined on a compact metric space that are uniformly bounded: $|g_n(x)| \leq M$ $\forall n$ and some $M \geq 0$, then, $\exists \{g_{n_k}\}$ s.t. $g_{n_k} \longrightarrow g\in{C(X)}$ uniformly.
$([0,1],|\cdot | )$ is a compact metric space...check. So, if we can show that $\{g_n\}$ is uniformly bounded and equicontinuous, then the Arzel{\'a}-Ascoli theorem will apply and give us the desired conclusion.
We now want to show that $\{g_n\}$ is uniformly bounded: By assumption, $f_n\in{L^2[0,1]}$. Let $h=1$ on $[0,1]$. Then H{\"o}lder's inequality with p=q=2 (Cauchy-Schwarz) gives us that $$||f_n \cdot 1||_1 \leq ||f_n||_2 \cdot ||1||_2$$ $$\int_0^1|f_n \cdot 1| \leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}} \left(\int_0^1|1^2|\right)^{\frac{1}{2}} = \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}$$ Squaring both sides, $$\left(\int_0^1|f_n|\right)^2 \leq \int_0^1|f_n|^2 \leq M$$ Thus, $f\in{L^1[0,1]}$. Note that we actually showed that $L^p[a,b] \subseteq L^1[a,b] \quad \forall 1 \leq p < \infty$. To compare these inequalities with the $g_n$ terms, $$|g_n|^2=\left|\int_0^xf_n\right|^2 \leq \left(\int_0^x|f_n|\right)^2 \leq \left(\int_0^1|f_n|\right)^2 \leq M$$ $$|g_n|^2 \leq M \Longrightarrow |g_n| \leq \sqrt{M}$$ Thus, $\{g_n\}$ is a uniformly bounded sequence...check.
Now consider $0\leq x < z \leq 1$. $$|g_n(x)-g_n(z)|=\left|\int_0^xf_n - \int_0^z f_n \right|=\left|\int_x^zf_n\right|\leq \int_x^z|f_n \cdot 1|$$ Using H{\"o}lder's inequality again, $$\leq \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\left(\int_x^z|1|^2\right)^{\frac{1}{2}} = \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|}\leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|} $$ $$ \leq \sqrt[4]{M}\sqrt{|x-z|}$$ So, $|g_n(x)-g_n(z)| \leq \sqrt[4]{M}\sqrt{|x-z|}$ which gives us a Lipschitz condition on $g_n$ $\forall n$. In the $\epsilon-\delta$ criterion for equicontinuity, if we choose $\delta \leq \frac{\epsilon^2}{\sqrt{M}}$ then $|g_n(x)-g_n(z)| \leq \epsilon$ making $\{g_n\}$ uniformly equicontinuous.
Friday, April 25, 2014
Royden 10.2.17
$\textbf{Problem 10.2.17}$ In a complete metric space, is the union of a countable collection of nowhere dense sets also nowhere dense?
$\textit{Solution}$: No. Consider $(\mathbb{R}, |\cdot |)$ which is complete as a metric space. A singleton point $\{x\} \in{\mathbb{R}}$ is nowhere dense. $closure\{x\}=\{x\}$ and $interior\{x\}= \emptyset$ since $\not \exists B_r(x) \subseteq \{x\}$. (Every ball about $x$ contains $x$, not the other way around). $\mathbb{Q}$ is countable, so we can write $ \mathbb{Q}= \cup_{i=1}^\infty \{q_i\}$ while $closure(\mathbb{Q})=\mathbb{R}$ and $interior(\mathbb{R})=\mathbb{R} \neq \emptyset$ ($\mathbb{R}$ is both open and closed). So, without additional conditions, we cannot determine if the countable union of nowhere dense subsets will be nowhere dense, dense, or neither.
$\textit{Solution}$: No. Consider $(\mathbb{R}, |\cdot |)$ which is complete as a metric space. A singleton point $\{x\} \in{\mathbb{R}}$ is nowhere dense. $closure\{x\}=\{x\}$ and $interior\{x\}= \emptyset$ since $\not \exists B_r(x) \subseteq \{x\}$. (Every ball about $x$ contains $x$, not the other way around). $\mathbb{Q}$ is countable, so we can write $ \mathbb{Q}= \cup_{i=1}^\infty \{q_i\}$ while $closure(\mathbb{Q})=\mathbb{R}$ and $interior(\mathbb{R})=\mathbb{R} \neq \emptyset$ ($\mathbb{R}$ is both open and closed). So, without additional conditions, we cannot determine if the countable union of nowhere dense subsets will be nowhere dense, dense, or neither.
Friday, April 11, 2014
Stein 2.21
$\textbf{Exercise 2.21}$ Let $f,g\in{L^1(\mathbb{R}^d)}$
(a) Prove that $f(x-y)g(y)$ is measurable in $\mathbb{R}^{2d}$
$\textit{Proof}$ : By proposition 2.3.7, with $g$ being measurable on $\mathbb{R}^d$, then the "sheet function" $\tilde{g}(x,y)=g(x)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$ . By proposition 2.3.9, with $f$ being measurable on $\mathbb{R}^d$, then the function $\tilde{f}(x,y)=f(x-y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$. With $f(x-y)g(y)$ being a product of 2 measurable functions and the previous result that the product of 2 measurable functions is measurable, we get that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2d}$. $\blacksquare$
Caution: In the context of Lebesgue measure, the product of 2 measurable functions is measurable. However, it is not true in general that the product of two integrable functions is integrable. Take $f(x)=\frac{1}{\sqrt{x}}\chi_{[0,1]}$. $\int |f| = \int_0^1 \frac{dx}{\sqrt{x}}=2$ however, $\int |f \cdot f|=\int_0^1\frac{dx}{x}=+ \infty$
To get any relationship between $L^p(E)$ and $L^q(E)$ for $p,q \in{[1, \infty]}$, we need $m(E)< \infty$. This follows from Hölder's inequality and is discussed on pages 142-3 in Royden.
(b) Show that if $f,g\in{L^1(\mathbb{R}^d)}$ then $f(x-y)g(y)\in{L^1(\mathbb{R}^{2d})}$
(c) Recall the definition of the convolution of $f$ and $g$ given by $$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy $$ Show that $f \ast g$ is well-defined for a.e. $x$ (that is, $f(x-y)g(y)\in{L^1(\mathbb{R}^{d})}$ for a.e. $x$)
(d) Show that $f \ast g$ is integrable whenever $f$ and $g$ are integrable and that $$||f \ast g||_{L^1} \leq ||f||_{L^1} \cdot ||g||_{L^1}$$ with equality when $f,g \geq 0$
$\textit{Proof}$ We combine the work of items (b)-(d) together.
$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy$. We want to show that $\int_{\mathbb{R}^d}|(f \ast g )(x)|dx < \infty $ i.e. $||f \ast g||_{L^1} < \infty $.
By the triangle inequality for Lebesgue integrals, $$\int_{\mathbb{R}^d}|(f \ast g )(x)|dx=\int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} f(x-y)g(y)dy \right| dx \leq \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)g(y)|dy \right) dx \quad (\star)$$ By the symmetry of $x$ and $y$ in Tonelli's theorem, $$=\int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dy \right) dx \overset{T}{=} \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dx \right) dy$$ By linearity, $$=\int_{\mathbb{R}^d} |g(y)| \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) dy= \left( \int_{\mathbb{R}^d} |g(y)| dy \right) \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) $$ Since the $L^1$ norm is translation invariant, $$=||g||_{L^1} \cdot ||f(x-y)||_{L^1}=||g||_{L^1} \cdot ||f||_{L^1}$$ We would get equality in $(\star)$ above if $f,g \geq 0$. $\blacksquare$
(e) The Fourier transform of an integrable function $f$ is defined by $$\hat{f}(\xi)=\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$$ Check that $\hat{f}$ is bounded and is a continuous function of $\xi$. Then, prove that $\forall \xi \in{\mathbb{C}}$ we get the identity: $$\widehat{(f \ast g)}(\xi)=\hat{f}(\xi)\hat{g}(\xi)$$ $\textit{Proof:}$ To check that $\hat{f}$ is bounded, we first observe that since $e^{-2\pi i x \xi}$ takes on points on the complex unit circle, so $|e^{-2\pi i x \xi}| \leq 1$ where $| \cdot |$ is the modulus (length in complex plane). Then, by the above observation and the triangle inequality, $$|\hat{f}(\xi)|=\left|\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx\right| \leq \int_{\mathbb{R}^d} |f(x)| \cdot |e^{-2\pi ix \xi}|dx \leq \int_{\mathbb{R}^d} |f(x)|dx = ||f||_{L^1} < \infty$$ So, $\hat{f}$ is bounded. To verify that $\hat{f}$ is a continuous function of $\xi$, we want to show that if $\xi_n \longrightarrow \xi$ then $\hat{f}(\xi_n) \longrightarrow \hat{f}(\xi) $ i.e. $\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi_n}dx \overset{\xi_n \rightarrow \xi}{\longrightarrow} \int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$ Label the integrands $\tilde{f}(x)=f(x)e^{-2 \pi ix \xi}$ and $\tilde{f}_n(x)=f(x)e^{-2 \pi ix \xi_n}$. Then, since $f\in{L^1(\mathbb{R}^d}) \Longrightarrow f\in{\mathcal{M}(\mathbb{R}^d})$ and so $\{\tilde{f}_n\}_{n=1}^\infty$ is a sequence of measurable functions. Also, $|\tilde{f}_n| \leq |f| \in{L^1(\mathbb{R}^d)}$. And, $\tilde{f}_n \longrightarrow \tilde{f}$ a.e. since $\lim_{n \rightarrow \infty } f(x)e^{-2 \pi ix \xi_n}=f(x) \lim_{n \rightarrow \infty} e^{-2 \pi i x \xi_n} = f(x)e^{-2 \pi i x \xi}$ by the continuity of the exponential function. By the Lebesgue dominated convergence theorem, $$\int_{\mathbb{R}^d} \tilde{f}_n(x)dx \longrightarrow \int_{\mathbb{R}^d} \tilde{f}_n(x)dx \quad i.e. \quad \hat{f}(\xi_n) \longrightarrow \hat{f}(\xi)$$
By definition, the Fourier transform of the convolution of $f$ and $g$ is $$(\widehat{f \ast g})(\xi)=\int_{\mathbb{R}^d}(f \ast g)(x)e^{-2\pi i x \xi} dx=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dy \right)e^{-2\pi i x \xi} dx$$ Using the symmetry of $x$ and $y$ in Fubini's theorem and the linearity of the integral, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dx \right)e^{-2\pi i x \xi} dy=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2\pi i x \xi}dx \right)g(y) dy$$ If we write $e^{-2 \pi i x \xi} = e^{-2 \pi i (x-y+y) \xi}=e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}$ then continuing our equality chain, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}dx \right)g(y) dy=\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)$$ By the translation invariance of the Lebesgue measure, $$=\left(\int_{\mathbb{R}^d}f(x)e^{-2 \pi i x \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)=\hat{f}(\xi)\hat{g}(\xi)$$
(a) Prove that $f(x-y)g(y)$ is measurable in $\mathbb{R}^{2d}$
$\textit{Proof}$ : By proposition 2.3.7, with $g$ being measurable on $\mathbb{R}^d$, then the "sheet function" $\tilde{g}(x,y)=g(x)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$ . By proposition 2.3.9, with $f$ being measurable on $\mathbb{R}^d$, then the function $\tilde{f}(x,y)=f(x-y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$. With $f(x-y)g(y)$ being a product of 2 measurable functions and the previous result that the product of 2 measurable functions is measurable, we get that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2d}$. $\blacksquare$
Caution: In the context of Lebesgue measure, the product of 2 measurable functions is measurable. However, it is not true in general that the product of two integrable functions is integrable. Take $f(x)=\frac{1}{\sqrt{x}}\chi_{[0,1]}$. $\int |f| = \int_0^1 \frac{dx}{\sqrt{x}}=2$ however, $\int |f \cdot f|=\int_0^1\frac{dx}{x}=+ \infty$
To get any relationship between $L^p(E)$ and $L^q(E)$ for $p,q \in{[1, \infty]}$, we need $m(E)< \infty$. This follows from Hölder's inequality and is discussed on pages 142-3 in Royden.
(b) Show that if $f,g\in{L^1(\mathbb{R}^d)}$ then $f(x-y)g(y)\in{L^1(\mathbb{R}^{2d})}$
(c) Recall the definition of the convolution of $f$ and $g$ given by $$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy $$ Show that $f \ast g$ is well-defined for a.e. $x$ (that is, $f(x-y)g(y)\in{L^1(\mathbb{R}^{d})}$ for a.e. $x$)
(d) Show that $f \ast g$ is integrable whenever $f$ and $g$ are integrable and that $$||f \ast g||_{L^1} \leq ||f||_{L^1} \cdot ||g||_{L^1}$$ with equality when $f,g \geq 0$
$\textit{Proof}$ We combine the work of items (b)-(d) together.
$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy$. We want to show that $\int_{\mathbb{R}^d}|(f \ast g )(x)|dx < \infty $ i.e. $||f \ast g||_{L^1} < \infty $.
By the triangle inequality for Lebesgue integrals, $$\int_{\mathbb{R}^d}|(f \ast g )(x)|dx=\int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} f(x-y)g(y)dy \right| dx \leq \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)g(y)|dy \right) dx \quad (\star)$$ By the symmetry of $x$ and $y$ in Tonelli's theorem, $$=\int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dy \right) dx \overset{T}{=} \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dx \right) dy$$ By linearity, $$=\int_{\mathbb{R}^d} |g(y)| \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) dy= \left( \int_{\mathbb{R}^d} |g(y)| dy \right) \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) $$ Since the $L^1$ norm is translation invariant, $$=||g||_{L^1} \cdot ||f(x-y)||_{L^1}=||g||_{L^1} \cdot ||f||_{L^1}$$ We would get equality in $(\star)$ above if $f,g \geq 0$. $\blacksquare$
(e) The Fourier transform of an integrable function $f$ is defined by $$\hat{f}(\xi)=\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$$ Check that $\hat{f}$ is bounded and is a continuous function of $\xi$. Then, prove that $\forall \xi \in{\mathbb{C}}$ we get the identity: $$\widehat{(f \ast g)}(\xi)=\hat{f}(\xi)\hat{g}(\xi)$$ $\textit{Proof:}$ To check that $\hat{f}$ is bounded, we first observe that since $e^{-2\pi i x \xi}$ takes on points on the complex unit circle, so $|e^{-2\pi i x \xi}| \leq 1$ where $| \cdot |$ is the modulus (length in complex plane). Then, by the above observation and the triangle inequality, $$|\hat{f}(\xi)|=\left|\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx\right| \leq \int_{\mathbb{R}^d} |f(x)| \cdot |e^{-2\pi ix \xi}|dx \leq \int_{\mathbb{R}^d} |f(x)|dx = ||f||_{L^1} < \infty$$ So, $\hat{f}$ is bounded. To verify that $\hat{f}$ is a continuous function of $\xi$, we want to show that if $\xi_n \longrightarrow \xi$ then $\hat{f}(\xi_n) \longrightarrow \hat{f}(\xi) $ i.e. $\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi_n}dx \overset{\xi_n \rightarrow \xi}{\longrightarrow} \int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$ Label the integrands $\tilde{f}(x)=f(x)e^{-2 \pi ix \xi}$ and $\tilde{f}_n(x)=f(x)e^{-2 \pi ix \xi_n}$. Then, since $f\in{L^1(\mathbb{R}^d}) \Longrightarrow f\in{\mathcal{M}(\mathbb{R}^d})$ and so $\{\tilde{f}_n\}_{n=1}^\infty$ is a sequence of measurable functions. Also, $|\tilde{f}_n| \leq |f| \in{L^1(\mathbb{R}^d)}$. And, $\tilde{f}_n \longrightarrow \tilde{f}$ a.e. since $\lim_{n \rightarrow \infty } f(x)e^{-2 \pi ix \xi_n}=f(x) \lim_{n \rightarrow \infty} e^{-2 \pi i x \xi_n} = f(x)e^{-2 \pi i x \xi}$ by the continuity of the exponential function. By the Lebesgue dominated convergence theorem, $$\int_{\mathbb{R}^d} \tilde{f}_n(x)dx \longrightarrow \int_{\mathbb{R}^d} \tilde{f}_n(x)dx \quad i.e. \quad \hat{f}(\xi_n) \longrightarrow \hat{f}(\xi)$$
By definition, the Fourier transform of the convolution of $f$ and $g$ is $$(\widehat{f \ast g})(\xi)=\int_{\mathbb{R}^d}(f \ast g)(x)e^{-2\pi i x \xi} dx=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dy \right)e^{-2\pi i x \xi} dx$$ Using the symmetry of $x$ and $y$ in Fubini's theorem and the linearity of the integral, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dx \right)e^{-2\pi i x \xi} dy=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2\pi i x \xi}dx \right)g(y) dy$$ If we write $e^{-2 \pi i x \xi} = e^{-2 \pi i (x-y+y) \xi}=e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}$ then continuing our equality chain, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}dx \right)g(y) dy=\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)$$ By the translation invariance of the Lebesgue measure, $$=\left(\int_{\mathbb{R}^d}f(x)e^{-2 \pi i x \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)=\hat{f}(\xi)\hat{g}(\xi)$$
Thursday, April 10, 2014
Royden 17.1.6 p. 341
\textbf{Exercise 17.1.6} Let $(X,\mathcal{M},\mu)$ be a measure space and let $X_0\in{\mathcal{M}}$. Define $\mathcal{M}_0$ to be the collection of sets in $\mathcal{M}$ that are subsets of $X_0$ and $\mu_0 = \mu|_{X_0}$. Show that $(X_0,\mathcal{M}_0,\mu_0)$ is a measure space.
This problem establishes that a "smaller" measure space can be defined within a measurable set.
\textit{Proof} : To show that $(X_0,\mathcal{M}_0,\mu_0)$ is a measure space, we need to show
(i) $\mathcal{M}_0 \neq \emptyset$
(ii) $\mathcal{M}_0$ is a $\sigma$-algebra (closed under compliments and countable unions and intersections)
(iii) $\mu_0 = \mu|_{X_0}$ is a measure.
(i) $\emptyset \in{\mathcal{M}}$ since $\mu(\emptyset)=0$. Also, $\emptyset \subseteq X_0$ since the empty set is always a subset of any set. Thus, $\emptyset$ is a measurable subset of $X_0$, thus $\emptyset \in{\mathcal{M}_0}$. So, $\mathcal{M}_0$ contains at least one set (the empty set), so is nonempty.
(ii) Let $E\in{\mathcal{M}_0}$. This means that $E\in{\mathcal{M}}$ and $E \subseteq X_0$. Since $\mathcal{M}$ is a $\sigma$-algebra, $E^c\in{\mathcal{M}}$. So, $E^c \cap X_0 \in{\mathcal{M}}$ (an intersection of 2 measurable sets). Also, $E^c \cap X_0$ is the relative compliment of $E$ in $X_0$. If we define our compliments in $\mathcal{M}_0$ in this way, then we get that the relative compliment of $E$ in $X_0$ is measurable and contained in $X_0$, thus in $\mathcal{M}_0$. This makes $\mathcal{M}_0$ closed under (relative) compliments in $X_0$. To show that $\mathcal{M}_0$ is closed under countable set unions, let $\{E_j\}_{j=1}^\infty$ be a sequence of sets s.t. $E_j \in{\mathcal{M}}$ and $E_j \subseteq X_0$ $\forall j\in{\mathbb{N}}$. $E_j \subseteq X_0 \quad \forall j \Longrightarrow \cup_{j=1}^\infty E_j \subseteq X_0$. $\mathcal{M}$ is a $\sigma$-algebra, so $\cup_{j=1}^\infty E_j \subseteq \mathcal{M}$. Thus, $\cup_{j=1}^\infty E_j \in{\mathcal{M}_0}$. DeMorgan's laws then automatically make $\mathcal{M}_0$ also closed under countable intersections. So, $\mathcal{M}_0$ is a $\sigma$-algebra.
(iii) $\mu$ is a measure on all of $\mathcal{M}$, so it is automatically a measure on any subset of $\mathcal{M}$. So, the restriction of $\mu$ to $\mathcal{M}_0$ is a measure.
$\therefore \quad (X_0,\mathcal{M}_0, \mu_0)$ is a measure space. $\blacksquare$
This problem establishes that a "smaller" measure space can be defined within a measurable set.
\textit{Proof} : To show that $(X_0,\mathcal{M}_0,\mu_0)$ is a measure space, we need to show
(i) $\mathcal{M}_0 \neq \emptyset$
(ii) $\mathcal{M}_0$ is a $\sigma$-algebra (closed under compliments and countable unions and intersections)
(iii) $\mu_0 = \mu|_{X_0}$ is a measure.
(i) $\emptyset \in{\mathcal{M}}$ since $\mu(\emptyset)=0$. Also, $\emptyset \subseteq X_0$ since the empty set is always a subset of any set. Thus, $\emptyset$ is a measurable subset of $X_0$, thus $\emptyset \in{\mathcal{M}_0}$. So, $\mathcal{M}_0$ contains at least one set (the empty set), so is nonempty.
(ii) Let $E\in{\mathcal{M}_0}$. This means that $E\in{\mathcal{M}}$ and $E \subseteq X_0$. Since $\mathcal{M}$ is a $\sigma$-algebra, $E^c\in{\mathcal{M}}$. So, $E^c \cap X_0 \in{\mathcal{M}}$ (an intersection of 2 measurable sets). Also, $E^c \cap X_0$ is the relative compliment of $E$ in $X_0$. If we define our compliments in $\mathcal{M}_0$ in this way, then we get that the relative compliment of $E$ in $X_0$ is measurable and contained in $X_0$, thus in $\mathcal{M}_0$. This makes $\mathcal{M}_0$ closed under (relative) compliments in $X_0$. To show that $\mathcal{M}_0$ is closed under countable set unions, let $\{E_j\}_{j=1}^\infty$ be a sequence of sets s.t. $E_j \in{\mathcal{M}}$ and $E_j \subseteq X_0$ $\forall j\in{\mathbb{N}}$. $E_j \subseteq X_0 \quad \forall j \Longrightarrow \cup_{j=1}^\infty E_j \subseteq X_0$. $\mathcal{M}$ is a $\sigma$-algebra, so $\cup_{j=1}^\infty E_j \subseteq \mathcal{M}$. Thus, $\cup_{j=1}^\infty E_j \in{\mathcal{M}_0}$. DeMorgan's laws then automatically make $\mathcal{M}_0$ also closed under countable intersections. So, $\mathcal{M}_0$ is a $\sigma$-algebra.
(iii) $\mu$ is a measure on all of $\mathcal{M}$, so it is automatically a measure on any subset of $\mathcal{M}$. So, the restriction of $\mu$ to $\mathcal{M}_0$ is a measure.
$\therefore \quad (X_0,\mathcal{M}_0, \mu_0)$ is a measure space. $\blacksquare$
Friday, April 4, 2014
Stein 6.10
$\textbf{Exercise 6.10}$ Suppose $\nu_1, \nu_2, \nu_3$ are signed measures and $\mu$ is a positive measure on $(X,\mathcal{M})$. Using the symbols $\perp$ (mutually singular) and $\ll$ (absolutely continuous) prove the following:
(a) If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $ then $\nu_1 + \nu_2 \perp \mu$
(b) If $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ then $\nu_1 + \nu_2 \ll \mu$
(c) If $\nu_1 \perp \nu_2$ then $|\nu_1| \perp |\nu_2|$
(d) $\nu \ll |\nu|$
(e) If $\nu \perp \mu $ and $\nu \ll \mu$ then $\nu=0$
$\textit{Proof} : $
(a) Assume that $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $. By the definition given in the book (there are several other equivalent definitions) $\nu_1 \perp \mu$ means that $\nu_1$ and $\mu$ are supported on disjoint subsets. That is, $\exists A,B \in{\mathcal{M}}$ where $A \cap B = \emptyset$ s.t. $$\nu_1(E)=\nu_1(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}} $$ $\nu_2 \perp \mu$ means that $\exists C,D \in{\mathcal{M}}$ where $C \cap D = \emptyset$ s.t. $$\nu_2(E)=\nu_2(E \cap C) \quad \text{and} \quad \mu(E)=\mu(E \cap D) \quad \forall E \in{\mathcal{M}} $$ To find disjoint sets of support for $\nu_1+\nu_2$ and $\mu$, we need to pick some combination of $A,B,C,D$ for our disjoint sets. Note that $(A \cup C) \cap (B \cap D)= \emptyset$.
$\mu(E \cap D)=\mu(E)$ means that the $\mu-$measure of any measurable set, intersected with $D$, is the $\mu$-measure of that set. Thus, $$\mu((B \cap D)\cap E) = \mu([E \cap B] \cap D)=\mu[E \cap B]=\mu(E) $$ Also, $$(\nu_1+\nu_2)((A \cup C)\cap E)=\nu_1((A \cup C)\cap E)+\nu_2((A \cup C)\cap E)$$ Since $\nu_1$ and $\nu_2$ are both measures, they are additive on disjoint sets. So, splitting $(A \cup C) \cap E$ into 2 disjoint subsets, we get $$\nu_1((A \cup C) \cap E)=\nu_1((E \cap A) \cup (E \cap (C \cap A^c)))$$ $$=\nu_1((E \cap A)) + \nu_1[E \cap (C \cap A^c)]$$ Since $\nu_1$ is supported on $A$, $$=\nu_1[E \cap (C \cap A^c)] + \nu_1([E \cap (C \cap A^c)] \cap A)= \nu_1(\emptyset)=0$$ Similarly, $$\nu_2((A \cup C) \cap E)=\nu_2((E \cap C) \cup (E \cap (A \cap C^c)))$$ $$=\nu_2(E \cap C) + \nu_2[E \cap (A \cap C^c)]=\nu_2(E)+ \nu_2([E \cap (A \cap C^c)] \cap C)=\nu_2(E)$$ So, we have exhibited two disjoint sets $A \cup C$ and $B \cap D$ s.t. $$(\nu_1+\nu_2)(E) = (\nu_1 + \nu_2)((A \cup C) \cap E) \quad \text{and} \quad \mu(E)=\mu((B \cap D) \cap E) \quad \forall E \in{\mathcal{M}} $$ Therefore, $(\nu_1+\nu_2) \perp \mu$
(b) Assume that $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ with $E \in{\mathcal{M}}$ and $\mu(E)=0$. Then, $(\nu_1+\nu_2)(E)=\nu_1(E)+\nu_2(E)=0+0=0$. Thus, $\nu_1+\nu_2 \ll \mu$
(c) Assume $\nu_1 \perp \nu_2$. By definition, $\exists A,B \in{\mathcal{M}}$ s.t. $A \cap B = \emptyset$ with $$\nu_1(E) = \nu_1(E \cap A) \quad \text{and} \quad \nu_2(E) = \nu_2(E \cap B) \quad \forall E\in{\mathcal{M}}$$ The variation of a measure is itself a measure and is defined as: $$|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|$$ where the supremum is taken over all countable, measurable partitions of $E$. With $\nu_1$ supported on $A$, we get $|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_1(E_j \cap A)|=|\nu_1|(E \cap A)$. Similarly, $|\nu_2|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_2(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_2(E_j \cap B)|=|\nu_2|(E \cap B)$. Thus, $|\nu_1| \perp |\nu_2|$
(d) With the added assumption that $\nu$ is $\sigma$-finite, we get that $$-|\nu| \leq \nu \leq |\nu|$$ So if $|\nu|(E)=0$ this immediately forces $\nu(E) = 0$. \\ Even without this condition, assuming $|\nu|(E) \equiv \sup \sum_{j =1}^{\infty}|\nu(E_j)|=0$ (where $E_j$'s form a measurable, countable partition of $E$) this implies $|\nu(E_j)|=0 \quad \forall j\in{\mathbb{N}}$. In particular, $E \subseteq E$ is a trivial partition of itself. Thus, $\nu(E)=0$. Therefore $|\nu|(E)=0 \Longrightarrow \nu(E)=0$ i.e. $\nu \ll |\nu|$
(e) Assume $\nu \perp \mu$ and $\nu \ll \mu$. So, $\exists A,B \in{\mathcal{M}}$, $A \cap B = \emptyset$ s.t. $$\nu(E)=\nu(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}}$$ Since $\mu$ is supported on $B$, consider $$\mu[E \cap A] = \mu([E \cap A] \cap B)=\mu(\emptyset)=0$$ Since $\nu \ll \mu$ this implies $$\nu[E \cap A] = 0 = \nu(E)$$ i.e. $\nu(E)=0 \quad \forall E \in{\mathcal{M}}$ $\quad \blacksquare$
(a) If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $ then $\nu_1 + \nu_2 \perp \mu$
(b) If $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ then $\nu_1 + \nu_2 \ll \mu$
(c) If $\nu_1 \perp \nu_2$ then $|\nu_1| \perp |\nu_2|$
(d) $\nu \ll |\nu|$
(e) If $\nu \perp \mu $ and $\nu \ll \mu$ then $\nu=0$
$\textit{Proof} : $
(a) Assume that $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $. By the definition given in the book (there are several other equivalent definitions) $\nu_1 \perp \mu$ means that $\nu_1$ and $\mu$ are supported on disjoint subsets. That is, $\exists A,B \in{\mathcal{M}}$ where $A \cap B = \emptyset$ s.t. $$\nu_1(E)=\nu_1(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}} $$ $\nu_2 \perp \mu$ means that $\exists C,D \in{\mathcal{M}}$ where $C \cap D = \emptyset$ s.t. $$\nu_2(E)=\nu_2(E \cap C) \quad \text{and} \quad \mu(E)=\mu(E \cap D) \quad \forall E \in{\mathcal{M}} $$ To find disjoint sets of support for $\nu_1+\nu_2$ and $\mu$, we need to pick some combination of $A,B,C,D$ for our disjoint sets. Note that $(A \cup C) \cap (B \cap D)= \emptyset$.
$\mu(E \cap D)=\mu(E)$ means that the $\mu-$measure of any measurable set, intersected with $D$, is the $\mu$-measure of that set. Thus, $$\mu((B \cap D)\cap E) = \mu([E \cap B] \cap D)=\mu[E \cap B]=\mu(E) $$ Also, $$(\nu_1+\nu_2)((A \cup C)\cap E)=\nu_1((A \cup C)\cap E)+\nu_2((A \cup C)\cap E)$$ Since $\nu_1$ and $\nu_2$ are both measures, they are additive on disjoint sets. So, splitting $(A \cup C) \cap E$ into 2 disjoint subsets, we get $$\nu_1((A \cup C) \cap E)=\nu_1((E \cap A) \cup (E \cap (C \cap A^c)))$$ $$=\nu_1((E \cap A)) + \nu_1[E \cap (C \cap A^c)]$$ Since $\nu_1$ is supported on $A$, $$=\nu_1[E \cap (C \cap A^c)] + \nu_1([E \cap (C \cap A^c)] \cap A)= \nu_1(\emptyset)=0$$ Similarly, $$\nu_2((A \cup C) \cap E)=\nu_2((E \cap C) \cup (E \cap (A \cap C^c)))$$ $$=\nu_2(E \cap C) + \nu_2[E \cap (A \cap C^c)]=\nu_2(E)+ \nu_2([E \cap (A \cap C^c)] \cap C)=\nu_2(E)$$ So, we have exhibited two disjoint sets $A \cup C$ and $B \cap D$ s.t. $$(\nu_1+\nu_2)(E) = (\nu_1 + \nu_2)((A \cup C) \cap E) \quad \text{and} \quad \mu(E)=\mu((B \cap D) \cap E) \quad \forall E \in{\mathcal{M}} $$ Therefore, $(\nu_1+\nu_2) \perp \mu$
(b) Assume that $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ with $E \in{\mathcal{M}}$ and $\mu(E)=0$. Then, $(\nu_1+\nu_2)(E)=\nu_1(E)+\nu_2(E)=0+0=0$. Thus, $\nu_1+\nu_2 \ll \mu$
(c) Assume $\nu_1 \perp \nu_2$. By definition, $\exists A,B \in{\mathcal{M}}$ s.t. $A \cap B = \emptyset$ with $$\nu_1(E) = \nu_1(E \cap A) \quad \text{and} \quad \nu_2(E) = \nu_2(E \cap B) \quad \forall E\in{\mathcal{M}}$$ The variation of a measure is itself a measure and is defined as: $$|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|$$ where the supremum is taken over all countable, measurable partitions of $E$. With $\nu_1$ supported on $A$, we get $|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_1(E_j \cap A)|=|\nu_1|(E \cap A)$. Similarly, $|\nu_2|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_2(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_2(E_j \cap B)|=|\nu_2|(E \cap B)$. Thus, $|\nu_1| \perp |\nu_2|$
(d) With the added assumption that $\nu$ is $\sigma$-finite, we get that $$-|\nu| \leq \nu \leq |\nu|$$ So if $|\nu|(E)=0$ this immediately forces $\nu(E) = 0$. \\ Even without this condition, assuming $|\nu|(E) \equiv \sup \sum_{j =1}^{\infty}|\nu(E_j)|=0$ (where $E_j$'s form a measurable, countable partition of $E$) this implies $|\nu(E_j)|=0 \quad \forall j\in{\mathbb{N}}$. In particular, $E \subseteq E$ is a trivial partition of itself. Thus, $\nu(E)=0$. Therefore $|\nu|(E)=0 \Longrightarrow \nu(E)=0$ i.e. $\nu \ll |\nu|$
(e) Assume $\nu \perp \mu$ and $\nu \ll \mu$. So, $\exists A,B \in{\mathcal{M}}$, $A \cap B = \emptyset$ s.t. $$\nu(E)=\nu(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}}$$ Since $\mu$ is supported on $B$, consider $$\mu[E \cap A] = \mu([E \cap A] \cap B)=\mu(\emptyset)=0$$ Since $\nu \ll \mu$ this implies $$\nu[E \cap A] = 0 = \nu(E)$$ i.e. $\nu(E)=0 \quad \forall E \in{\mathcal{M}}$ $\quad \blacksquare$
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