Problem 1: Let $T$ be a compact, linear operator on a Hilbert Space $H$ and let $\lambda \neq 0$.
(i) Show that the range of $\lambda I - T$,
$$ \{g \in{H} : \exists f\in{H} \quad s.t. \quad (\lambda I - T)f = g \} $$
is a closed subspace of $H$.
\textit{Solution}: In the proof of Lax-Milgram, the coercivity condition of the bilinear functional $\phi$ allowed us to develop the inequality
$$||x|| \leq \frac{1}{K} ||Ax|| \quad \text{for some $K > 0$}$$
This allowed us to show that $Ran(A)$ was closed. Let $E_\lambda = Null(\lambda I - T)$. Since $E_\lambda$ is closed, we have the decomposition $H = E_\lambda \oplus E_\lambda^\perp$. Then,
$$\lambda I-T = (\lambda I -T)|_{E_\lambda}+(\lambda I-T)|_{E_\lambda^\perp} = 0|_{E_\lambda}+S$$
If we can show that
$$\exists c>0 \quad \text{s.t.} \quad c||f|| \leq ||Sf|| \quad \forall f\in{E_\lambda^\perp} \quad (1)$$
then taking $g_n \in{Ran(\lambda I-T)}$ and $g_n \longrightarrow g$ we get $g_n = 0|_{E_\lambda}(f_n) + S(f_n) \longrightarrow g$. So, $\{Sf_n\}$ converges, implying $\{Sf_n\}$ is Cauchy:
$$||Sf_n-Sf_m|| \longrightarrow 0 \quad \text{as} \quad n,m \longrightarrow \infty $$
By (1) and the linearity of $S$,
$$c||f_n-f_m|| \leq ||Sf_n-Sf_m|| \longrightarrow 0$$
So, $\{f_n\}$ is Cauchy. Since $H$ is complete, $\{f_n\}$ is convergent: $f_n \longrightarrow f$. Since $S$ is continuous, $Sf_n \longrightarrow Sf$. But, $Sf_n \longrightarrow g$ as well. So, $Sf = g$ i.e. $g\in{Ran(S)} = Ran(\lambda I-T)$.
\textit{Proof of (1):} Suppose to the contrary that $\forall M>0$, $\exists h_M \in{E_\lambda^\perp}$ s.t. $M||h|| > ||Sh||$. So, $M>||Su||$ for some $||u||=1$. Equivalently, $\exists \{u_n\} \subseteq E_\lambda^\perp$, $||u_n||=1$ s.t. $||Su_n||\longrightarrow 0$ i.e. $$||(\lambda I-T)u_n|| \longrightarrow 0$$
Since $E_\lambda \cap E_\lambda^\perp =\{0\}$ and $||u_n||=1$, $\{u_n\}$ cannot converge to an element in $E_\lambda$.
In other words, $\lambda$ is an approximate eigenvalue of $T|_{E_\lambda^\perp}$. Since $\lambda \neq 0$ and $T|_{E_\lambda^\perp}$ being the restriction of a compact operator is compact, $\lambda$ is an eigenvalue of $T|_{E_\lambda^\perp}$...which is a contradiction since all eigenvectors with eigenvalue $\lambda$ are contained in $E_\lambda$. $\square$
(ii) Show by example that $Ran(\lambda I-T)$ may not be closed if $\lambda = 0$.
\textit{Solution:} With $\lambda = 0$, we want to find an example of a linear, compact operator $T$ s.t. $Ran(T)$ is not a closed subset of $H$. Consider
$$H = l_2(\mathbb{N}) = \{(x_j):x_j \in{\mathbb{C}}, \sum_{j=1}^\infty|x_j|^2 < \infty \}$$
and let $$T(x_1,x_2,x_3,...) = (x_1,\frac{x_2}{2},...,\frac{x_n}{n},...)$$
Then, $T$ is compact since $T$ is the limit of a sequence of finite rank operators: $T_1(x_1,x_2,...) = (x_1,0,0,...)$, $T_2 = (x_1,\frac{x_2}{2},0,0,...)$ etc. and
$(T-T_n)x = (0,0,...,0,\frac{x_{n+1}}{n+1},\frac{x_{n+2}}{n+2},...)$. So,
$$||T-T_n|| = \sup_{||u||=1}||(T_n-T)u|| \leq \frac{1}{n} \longrightarrow 0$$
$T$ also happens to be symmetric since $\left\langle Tx,y \right\rangle=\sum_{j=1}^\infty \frac{x_j}{j}y_j=\sum_{j=1}^\infty x_j\frac{y_j}{j}= \left\langle x, Ty \right\rangle$. Now, $x=\{\frac{1}{j}\}_{j=1}^\infty \in{l_2}$ however, $y=\{1\}$ is s.t. $Ty=x$ yet $y \not \in{l_2}$. So, $Ran(T)$ is not all of $l_2$. However, given any $y=(y_1,y_2,...) \in{l_2}$ then $x^{(n)} = (y_1,2y_2,...,ny_n,0,0,...)$ is s.t. $Tx_n \longrightarrow y$. So, $\overline{Ran(T)}=l_2$.
(iii) Show that $\lambda I-T$ is invertible iff $\bar{\lambda} I -T^*$ is invertible.
\textit{Solution:} We will show the following circle of implications: $Ran(\lambda I-T)=H \Longrightarrow \bar{\lambda}I-T^*$ is invertible $\Longrightarrow \lambda I - T$ is invertible.
Assume that $\lambda I-T$ is surjecive i.e. $Ran(\lambda I -T)=H$. Let $h\in{Null(\bar{\lambda}I-T^*)}$ and let $f\in{H}$ be arbitrary. Since $Ran(\lambda I_T)=H$, there exists $g\in{H}$ s.t. $(\lambda I-T)g=f$. Then, $\left\langle f,h \right\rangle=\left\langle (\lambda I-T)g,h \right\rangle=\left\langle g,(\lambda I-T)^*h \right\rangle = \left\langle g,(\bar{\lambda} I-T^*)h \right\rangle = \left\langle g,0 \right\rangle = 0$. Since $f$ was arbitrary, this implies that $h=0$ i.e. $Null(\bar{\lambda}I-T^*) = \{0\}$. Thus, $\bar{\lambda} I-T^*$ is invertible.
Now let $S = \bar{\lambda} I-T^*$. We want to show that $S^*$ is invertible by showing that $S^*(S^{-1})^*=(S^{-1})^*S^* = I$. Let $f,g\in{H}$. Since $S^{-1}$ exists and is bounded,
$\left\langle f,(S^{-1})^*S^*g \right\rangle = \left\langle S^{-1}f,S^*g \right\rangle = \left\langle SS^{-1}f,g \right\rangle = \left\langle f,g \right\rangle = \left\langle S^{-1}Sf,g \right\rangle = \left\langle Sf,(S^{-1})^*g \right\rangle = \left\langle f,S^*(S^{-1})^*g \right\rangle$. Therefore, $S^* = \lambda I-T$ is invertible and, in particular, surjective. This completes the circle of implications.
Saturday, April 9, 2016
Saturday, July 19, 2014
Beppo Levi's Lemma
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. $\blacksquare$
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. $\blacksquare$
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 $\mu(\limsup E_k) = 0$ $\blacksquare$
Subscribe to:
Comments (Atom)