HW8

Exercise 9

22. Let $f\in L^p\cap L^{\infty }$ for some $p<\infty$, so that $f\in L^q$ for all $q>p$. Prove: $\Vert f{\Vert }_{\infty }={\lim }_{q\to \infty }\Vert f{\Vert }_q$.

\begin{proof} For any simple function $g={\sum }_{i=1}^n{a}_i{\chi }_{E_i}$, we have $||g|{|}_q=({\sum }_{k=1}^n\mu (E_i)a_i^q{)}^{1/q}$ and $||g|{|}_q\to {\max }_i{a}_i=||g|{|}_{\infty }$. Since $f$ is measurable, there exists a sequence of simple functions $\{g_k\}$ such that $|g_k|\uparrow |f|$. Then by MCT we have ${\lim }_q||f|{|}_q=||f|{|}_{\infty }$. \end{proof}

Exercise 10

8. For $A\in \mathcal{L}\left({\mathbb{R}}^n\right)$, let $f_A(x)=|A\cap B(0,|x|)|$. Find ${\lim }_{|x|\to \infty }{f}_A(x)$ and show that $f_A$ is continuous on ${\mathbb{R}}^n$.

\begin{proof} Note that $|A|=|A\cap B(0,|x|)|\bigsqcup |A\cap B(0,|x|{)}^c|$. When $|x|\to \infty$, $|B(0,|x|{)}^c|\to 0$ and so $f_A(x)\to |A|$. Therefore, ${\lim }_{|x|\to \infty }{f}_A(x)=|A|$.

It remains to show $f_A$ is continuous on ${\mathbb{R}}^n$. For any $ϵ>0$ and a given $x\in {\mathbb{R}}^n$, there exists $\delta$ such that $|B(0,|x|+\delta )|-|B(0,|x|)|<ϵ$. Then $|f_A(x+y)-f_A(x)|<ϵ$ for all $y\in B(x,\delta )$ and so $f_A$ is continuous on $x$. By the arbitrary of $x$, $f_A$ is continuous on ${\mathbb{R}}^n$. \end{proof}

12. Let $E$ be a given subset of ${\mathbb{R}}^n$. Show that the following five statements are equivalent:

• (a) $E$ is Lebesgue measurable;
• (b) For every $ϵ>0$, there exists an open set $O\supset E$ such that ${\mu }^{\ast }(O-E)<ϵ$;
• (c) There exists a $G_{\delta }$-set $G\supset E$ such that ${\mu }^{\ast }(G-E)=0$;
• (d) For every $ϵ>0$, there exists a closed set $F\subset E$ such that ${\mu }^{\ast }(E-F)<ϵ$;
• (e) There exists an $F_{\sigma }$-set $F\subset E$ such that ${\mu }^{\ast }(E-F)=0$.

\begin{proof} a)→b) Since $E$ is Lebesgue measurable, by the regularity of Lebesgue outer measure there exists open $U\supseteq E$ such that ${\mu }^{\ast }(U)-ϵ⩽{\mu }^{\ast }(E)⩽{\mu }^{\ast }(U)$. Since $E$ is measurable, we have ${\mu }^{\ast }(O)={\mu }^{\ast }(O\cap E)+{\mu }^{\ast }(O\cap E^c)$ and so ${\mu }^{\ast }(O-E)<ϵ$.

b)→c) Define open $U_n$ such that ${\mu }^{\ast }(U_n-E)<1/n$ and $U_n\supseteq E$. Define $G={\cap }_{n=1}^{\infty }{U}_n$, then $G$ is a $G_{\delta }$-set and ${\mu }^{\ast }(G-E)⩽{\mu }^{\ast }(U_n-E)<1/n$ for all $n\in {\mathbb{N}}_{+}$. Therefore, ${\mu }^{\ast }(G-E)=0$.

c)→d) Since $E$ is Lebesgue measurable, $E^c$ is also lebesgue measurable. Then there exists $G_{\delta }$-set $G:={\cap }_{n=1}^{\infty }{G}_n$ such that ${\mu }^{\ast }(G-E^c)=0$. It deduces that ${\mu }^{\ast }(E-G^c)={\mu }^{\ast }(G-E^c)=0$. Note that $G^c={\cup }_{n=1}^{\infty }{G}_n^c$ and $F_m:={\cup }_{n=1}^m{G}_n^c$ satisfies $F_m\uparrow G^c$ and $F_m$ closed. Hence $(E-F_m)\downarrow (E-G^c)$ with null set $E-G^c$. Therefore, for any $ϵ>0$, there exists $m$ such that ${\mu }^{\ast }(E-F_m)<ϵ$.

d)→e) Define closed $F_n$ such that ${\mu }^{\ast }(E-F_n)<1/n$ and $F_n\subseteq E$. Define $F={\cup }_{n=1}^{\infty }{F}_n$, then $F$ is a $F_{\sigma }$-set and ${\mu }^{\ast }(E-F)⩽{\mu }^{\ast }(E-F_n)<1/n$ for all $n\in {\mathbb{N}}_{+}$. Therefore, ${\mu }^{\ast }(E-F)=0$.

e)→a) Since $F\in \mathcal{B}({\mathbb{R}}^n)$, $F$ is Lebesgue measurable. It deduces that

$${\mu }^{\ast }(A\cap E)={\mu }^{\ast }(A\cap E\cap F)+{\mu }^{\ast }(A\cap E\cap F^c)={\mu }^{\ast }(A\cap E\cap F)$$

for any subset $A\subseteq {\mathbb{R}}^n$. For any given $A\subseteq {\mathbb{R}}^n$, we have

$${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap F)+{\mu }^{\ast }(A\cap F^c)⩾{\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c)$$

as $A\cap F^c\supseteq A\cap E^c$. Therefore, $E$ is Lebesgue measurable. \end{proof}

Exercise 11

1. Let $\mathcal{M}$ be an algebra of subsets of $X,\mathcal{N}$ an algebra of subsets of $Y$, and $\mathcal{R}=\{A\times B\subset X\times Y:A\in \mathcal{M},B\in \mathcal{N}\}$ the rectangles in $\mathcal{M}\times \mathcal{N}$. Prove that the collection of finite unions of rectangles in $\mathcal{M}\times \mathcal{N}$ form an algebra of subsets of $X\times Y$.

\begin{proof} Define $\mathcal{A}$ as the set of finite unions of rectangles in $\mathcal{M}\times \mathcal{N}$. Note that $\varnothing ,X\times Y\in \mathcal{A}$. It is easy to show for any $\{A_i{\}}_{i=1}^n\subseteq \mathcal{A}$, there is ${\cup }_{i=1}^n{A}_i\in \mathcal{A}$. Assume that $A={\cup }_{k=1}^m{A}_k\times B_k\in \mathcal{A}$ where $A_k\in X$ and $B_k\in X$. Then

$$A^c={\cap }_{k=1}^m(A_k\times B_k{)}^c={\cap }_{k=1}^m((A_k\times B_k^c)\cup (A_k^c\times B_k)\cup (A_k^c\times B_k^c)).$$

Note that $(S_1\cup S_2)\cap (K_1\cup K_2)=(S_1\cap K_1)\cup (S_2\cap K_2)\cup (S_1\cap K_1)\cup (S_2\cap K_2)$ and for any rectangles $A\times B$ and $C\times D$, its intersection $(A\times B)\cap (C\times D)=(A\cap C)\times (B\times D)$ is still a rectangle. Then it is easy to show $A^c$ is an element of $\mathcal{A}$. Therefore, $\mathcal{A}$ is an algebra. \end{proof}

2. Show that $\mathcal{B}\left({\mathbb{R}}^n\right)\otimes \mathcal{B}\left({\mathbb{R}}^m\right)=\mathcal{B}\left({\mathbb{R}}^{n+m}\right)$.

\begin{proof} Recall that $\mathcal{B}({\mathbb{R}}^n)$ is the $\sigma$-algebra generated by sets in the form of $\{(x_1,\cdots ,x_n):x_i⩽b_i,i=1,\cdots ,n,b_i\in \mathbb{R}\}$. Notice that $\mathcal{B}({\mathbb{R}}^n)\otimes \mathcal{B}({\mathbb{R}}^m)$ contains element in the form of

$$\begin{array}{rl}& \{(x_1,\cdots ,x_n):x_i⩽b_i,i=1,\cdots ,n,b_i\in \mathbb{R}\}\times \{(x_{n+1},\cdots ,x_{n+m}):x_i⩽b_i,i=n+1,\cdots ,m+n,b_i\in \mathbb{R}\}\\ =& \{(x_1,\cdots ,x_{m+n}):x_i⩽b_i,i=1,\cdots ,m+n,b_i\in \mathbb{R}\}\end{array}$$

for any $b_1,\cdots ,b_{m+n}$. Since LHS is an $\sigma$-algebra, we have $\mathcal{B}\left({\mathbb{R}}^n\right)\otimes \mathcal{B}\left({\mathbb{R}}^m\right)\supseteq \mathcal{B}\left({\mathbb{R}}^{n+m}\right)$. On the other hand, for any $A\in \mathcal{B}({\mathbb{R}}^n)$ and $B\in \mathcal{B}({\mathbb{R}}^m)$, $A\times B\in {\mathbb{R}}^{m+n}$. Since LHS is generated by measurable rectangles and $\mathcal{B}({\mathbb{R}}^{m+n})$ is an $\sigma$-algebra, we have $\mathcal{B}\left({\mathbb{R}}^n\right)\otimes \mathcal{B}\left({\mathbb{R}}^m\right)\subseteq \mathcal{B}\left({\mathbb{R}}^{n+m}\right)$. \end{proof}

Exercise 12

1. Show that if $f\in L^1\left({\mathbb{R}}^n\right),f\ne 0$, then there exist $C,R>0$ such that $f^{\ast }(x)>C|x{|}^{-n}$ for $|x|>R$.

\begin{proof} Since simple measurable functions determine a dense subspace of $L^p$, there exists a simple function $\varphi$ with bounded support $\Omega$ such that $||f-\varphi |{|}_1<ϵ$ and so

$$ϵ>{\int }_{{\mathbb{R}}^n}|f(x)-\varphi (x)|dx>{\int }_{{\Omega }^c}|f(x)|dx.$$

It deduces that ${\int }_{\Omega }|f(x)|dx>||f|{|}_1-ϵ$.

Take $R$ such that $B(0,R)\supseteq \Omega$. For any $x\in {\mathbb{R}}^n$ with $|x|>R$, we have $B(x,2|x|)\supseteq \Omega$ and

$$f^{\ast }(x)⩾\frac{1}{m(B(x,2|x|))}{\int }_{B(x,2|x|)}|f(x)|dx⩾\frac{1}{m(B(x,2|x|))}{\int }_{\Omega }|f(x)|dx>\frac{1}{m(B(x,2|x|))}(||f|{|}_1-ϵ)=C|x{|}^{-n}$$

where $C=(||f|{|}_1-ϵ)|x{|}^n/m(B(x,2|x|))=(||f|{|}_1-ϵ)/(2^{n}m(B(0,1)))$ is a constant. \end{proof}

2. Define

$$H^{\ast }f(x)=\sup \left\{\frac{1}{|B|}{\int }_B|f(y)|dy:B\text{ is a ball and }x\in B\right\}.$$

Show that $f^{\ast }\le H^{\ast }f\le 2^n{f}^{\ast }$.

\begin{proof} Define $S_1:=\{B(x,r):r>0\}$. Then $S_1\subseteq \{B:B\text{mbox}isaballandx\in B\}$. Then $f^{\ast }(x)=\sup \frac{1}{|B|}{\int }_B|f(y)|dy$ with $B\in S_1$ and so $f^{\ast }⩽H^{\ast }f$.

For any ball $B$ with radius $r$ and $x\in B$, define $B_1=B(x,2r)$ and then $B_1\supseteq B$. It follows that

$$\frac{1}{|B|}{\int }_B|f(y)|dy⩽\frac{1}{|B|}{\int }_{B_1}|f(y)|dy=\frac{2^n}{|B_1|}{\int }_{B_1}|f(y)|dy⩽2^n{f}^{\ast }.$$

By the arbitrary of $B$, we have $H^{\ast }f⩽2^n{f}^{\ast }$. \end{proof}