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, $\mathcal{M} \supseteq \mathcal{M}'$.
$\therefore \mathcal{M}=\mathcal{M}'$ $\blacksquare$
Monday, June 23, 2014
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, $$m(\delta E) = m_*(\delta E) = \inf \sum_{j=1}^\infty |\delta Q_j| = \delta_1 \cdots \delta_d \cdot \inf \sum_{j=1}^\infty |Q_j|=\delta_1 \cdots \delta_d \cdot m_*(E) = \delta_1 \cdots \delta_d \cdot m(E)$$
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