HW4

Exercise 5

7. Let $T:X\to Y$ be a map. Show that $T^{-1}(\sigma (G))=\sigma \left(T^{-1}(G)\right)$ holds for any families $G$ of subsets of $Y$.

\begin{proof} By homework 1 exercise 9, for any $\sigma$-algebra on $Y$, $T^{-1}(Y)$ is also a $\sigma$-algebra. Thus, it follows that $T^{-1}(\sigma (G))\supseteq \sigma (T^{-1}(G))$ as $T^{-1}(G)\subseteq T^{-1}(\sigma (G))$.

On the other hand, it is easy to show that $T({\cup }_{i=1}^{\infty }{T}^{-1}(A_i))={\cup }_{i=1}^{\infty }{T}^{-1}(A_i)$ and $T(T^{-1}(A^c))=(T(T^{-1}(A)){)}^c$. Thus, for any $A\in \sigma (T^{-1}(G))$, we have that $T(A)\in \sigma (G)$ and $A\in T^{-1}(\sigma (G))$. Therefore, $T^{-1}(\sigma (G))\supseteq \sigma (T^{-1}(G))$ and so $T^{-1}(\sigma (G))=\sigma (T^{-1}(G))$. \end{proof}

8. Let $(X,\mathcal{A})$ be a measurable space.

• (i) Let $f,g:X\to \mathbb{R}$ be measurable functions. Show that for every $A\in \mathcal{A}$ the function $h(x):=f(x)$, if $x\in A$, and $h(x):=g(x)$, if $x\notin A$, is measurable.
• (ii) Let $\left\{f_n\right\}$ be a sequence of measurable functions and let $\left\{A_n\right\}\subset \mathcal{A}$ such that ${\cup }_n{A}_n=X$. Suppose that ${f_n|}_{A_n\cap A_k}={f_k|}_{A_n\cap A_k}$ for all $k,n\in \mathbb{N}$ and set $f(x):=f_n(x)$ if $x\in A_n$. Show that $f:X\to \mathbb{R}$ is measurable.

\begin{proof} i) For any open subset $U\subseteq \mathbb{R}$, $h^{-1}(U)=(f^{-1}(U)\cap A)\cup (g^{-1}(U)\cap A^c)$. Since $f,g$ are measurable functions, $f^{-1}(U)$ and $g^{-1}(U)$ are measurable. Note that $A,A^c$ are measurable, then $h^{-1}(U)$ is measurable and so $h$ is a measurable function.

ii) Since ${f_n|}_{A_n\cap A_k}={f_k|}_{A_n\cap A_k}$, $f$ is well-defined. Note that for any open subset $U\subseteq R$,

$$f^{-1}(U)=\bigcup _{n=1}^{\infty }(f_n^{-1}(U)\cap A_n)$$

where $f_n^{-1}(U)$ are measurable for all $n\in {\mathbb{N}}_{+}$. Therefore, $f^{-1}(U)$ is measurable. \end{proof}

9. Show that any left- or right-continuous function $u:\mathbb{R}\to \mathbb{R}$ is measurable.

\begin{proof} Assume that $u$ is left-continuous.

Firstly, we claim that $u$ has only countable discontinuities. Define $A:=\{x\in \mathbb{R}:u\text{mbox}isnotcontinuousonx\}$. Suppose for contradiction that $A$ is uncountable. For $i\in {\mathbb{Z}}^{+}$, let $A_i$ be the set of all $a\in A$ such that, for any neighborhood $U$ of $a$, $|f(x)-f(y)|⩾1/i$ holds for some $x,y\in U$. Since $f$ is discontinuous on $A$, we have $A={\bigcup }_{i=1}^{\infty }{A}_i$ and it follows that $A_n$ is uncountable for some $n$. Call a point $a\in A_n$ left-isolated if $(a-ϵ,a)$ does not meet $A_n$ for sufficiently small $ϵ>0$. By associating to each left isolated point $a$ a rational number from the interval ( $a-ϵ,a$ ), we see that $A_n$ has countably many left-isolated points. Since $A_n$ is uncountable, there exists some $p\in A_n$ which is not left isolated. For any $ϵ>0$, we have that $(p-ϵ,p)$ is a neighborhood of some $a\in A_n$. Hence, $|u(x)-u(y)|⩾1/n$ for some $x,y\in (p-ϵ,p)$, which contradicts the assumption that the limit as $x\to p$ from the left of $u(x)$ exists.

To show $u$ is continuous, it suffices to show $u^{-1}((a,\infty ])$ is measurable for every $a\in \mathbb{R}$. Assume that $u^{-1}((a,\infty ])=A$ and $\{a_i{\}}_{i\in I}\in \mathbb{R}\cup \{-\infty ,\infty \}$ is the set of countable discontinuities with $a_n<a_m$ if $n<m$ and $I\subseteq {\mathbb{N}}_{+}$. Since $u{|}_{(a_n,a_{n+1}]}$ is continuous for all $n,n+1\in I$, we have that $(u{|}_{(a_n,a_{n+1}]}{)}^{-1}((a,\infty ])=A\cap (a_n,a_{n+1}]$ is measurable. Note that

$$A={\cup }_{n,n+1\in I}((a_n,a_{n+1}]\cap A)$$

is measurable and so $u$ is measurable.

If $u$ is right-continuous, then $-u$ is left-continuous and it is measurable. It follows that $u$ is measurable. \end{proof}

10. Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ be continuous, 1-1, and onto. Prove that $f$ maps Borel sets onto Borel sets.

\begin{proof} Note that $f$ is Borel measurable by $f$ continuous. Let $\mathcal{B}$ be the Borel sets of ${\mathbb{R}}^n$. Then for any $B\in \mathcal{B}$, we have that $f^{-1}(B)\in \mathcal{B}$. Define

$$\phi :\mathcal{B}\to \mathcal{B},B\mapsto f^{-1}(B).$$

Assume that there exist distinct $B_1,B_2\in \mathcal{B}$ such that $\phi (B_1)=\phi (B_2)$, then it follows that $f^{-1}(B_1)=f^{-1}(B_2)$ and $f(f^{-1}(B_1))=f(f^{-1}(B_1))$, which is impossible. Therefore, $f$ is injective.

For each $B\in \mathcal{B}$, there exists $\phi (B)\in \mathcal{B}$ such that $f(\phi (B))=B$ and so $f$ maps Borel sets onto Borel sets. \end{proof}

Exercise 6

1. Suppose $f_n$ are measurable functions. Prove that

$$A=\left\{x:\underset{n\to \infty }{\lim }{f}_n(x)\text{ exists }\right\}$$

is a measurable set.

\begin{proof} Let $g(x)=\lim {\sup }_{n\to \infty }{f}_n(x)$ and let $h(x)=\lim {\inf }_{n\to \infty }{f}_n(x)$. Then $g$ and $h$ are measurable functions. Note that $A=\{x:g(x)=h(x)\}=(g-h{)}^{-1}(0)$ and $g-h$ is measurable, so it deduces that $A$ is measurable set. \end{proof}

2. Suppose $(X,\mathcal{A})$ is a measurable space, $f$ is a real-valued function, and $\{x:f(x)>r\}\in \mathcal{A}$ for each rational number $r$. Prove that $f$ is measurable.

\begin{proof} As $\{x:f(x)>r\}\in \mathcal{A}$ for all $r\in \mathbb{Q}$, we have that $f^{-1}((r,\infty ])$ is measurable for all $r\in \mathbb{Q}$. For any $s\in \mathbb{R}\setminus \mathbb{Q}$, there exists $\{r_n\}\subseteq \mathbb{Q}$ such that $r_n$ is a decreasing sequence and $r_n\to s,n\to \infty$. Define $A_n=\{x:f(x)>r_n\}$, then $\{A_n{\}}_{n=1}^{\infty }$ is a sequence of measurable sets and so ${\cup }_{n=1}^{\infty }{A}_n\in \mathcal{A}$. Note that

$${\cup }_{i=1}^{\infty }{A}_n=\{x:f(x)>s\}=f^{-1}((s,\infty ])$$

is measurable. Therefore, for every $a\in \mathbb{R}$, $f^{-1}((a,\infty ])$ is measurable and so $f$ is measurable. \end{proof}

4. Let $(X,\mathcal{A})$ be a measurable space. If $X=A\cup B$ where $A,B\in \mathcal{A}$. Show that a function $f$ on $X$ is measurable iff $f$ is measurable on $A$ and on $B$.

\begin{proof} Assume that $f$ is measurable on $X$ and $f:X\to Y$ with measurable space $(Y,\mathcal{C})$. We aim to show $f$ is measurable on $A$ and $B$. Note that for any $C\in \mathcal{C}$,

$$(f{|}_A{)}^{-1}(C)=f^{-1}(C)\cap A$$

is measurable. Therefore, $f$ is measurable on $A$. Similarly, we can prove that $f$ is measurable on $B$.

Now we assume that $f$ is measurable on $A$ and $B$. For any $C\in \mathcal{C}$, by $X=A\cup B$, we have that

$$f^{-1}(C)=\left((f{|}_A{)}^{-1}(C)\right)\cup \left((f{|}_b{)}^{-1}(C)\right).$$

Since $(f{|}_A{)}^{-1}(C)$ and $(f{|}_B{)}^{-1}(C)$ are measurable, $f^{-1}(C)$ is also measurable. Therefore, $f$ is measurable on $X$. \end{proof}

5. Let $(X,\mathcal{A},\mu )$ be a measure space and $\left\{f_n\right\},f$ measurable functions on $X$. Prove:

$$\left\{x\in X:\underset{n\to \infty }{\lim }{f}_n(x)\ne f(x)\right\}={\cup }_{k=1}^{\infty }{\cap }_{m=k}^{\infty }\left\{x:\underset{n>m}{\sup }\left|f_n(x)-f(x)\right|>1/k\right\}.$$

\begin{proof} For any $x\in {\cup }_{k=1}^{\infty }{\cap }_{m=k}^{\infty }\left\{x:{\sup }_{n>m}\left|f_n(x)-f(x)\right|>1/k\right\}$, there exists $k\in {\mathbb{N}}_{+}$ such that $x\in {\cap }_{m=k}^{\infty }\{x:{\sup }_{n>m}\left|f_n(x)-f(x)\right|>1/k\}$. Hence, for any $N\in {\mathbb{N}}_{+}$, there exists $n>N$ such that $|f_n(x)-f(x)|>1/k$ and so $f_n(x)\nrightarrow f(x)$ as $n\to \infty$. Therefore, $x\in \left\{x\in X:{\lim }_{n\to \infty }{f}_n(x)\ne f(x)\right\}$.

For any $x\in \left\{x\in X:{\lim }_{n\to \infty }{f}_n(x)\ne f(x)\right\}$, either $f(x)\ne {\mathrm{limsup}}_{n\to \infty }{f}_n(x)$ or $f(x)\ne {\mathrm{liminf}}_{n\to \infty }{f}_n(x)$ holds. If $f(x)\ne {\mathrm{limsup}}_{n\to \infty }{f}_n(x)$, then there exists $k\in {\mathbb{N}}_{+}$ such that

$$|f(x)-\underset{n\to \infty }{\mathrm{limsup}}{f}_n(x)|>\frac{1}{k}$$

and so for any $N$ there exists $n>N$ such that $|f_n(x)-f(x)|>\frac{1}{k}$. Therefore, $x\in {\cup }_{k=1}^{\infty }{\cap }_{m=k}^{\infty }\left\{x:{\sup }_{n>m}\left|f_n(x)-f(x)\right|>1/k\right\}$. Similarly, if $f(x)\ne {\mathrm{liminf}}_{n\to \infty }{f}_n(x)$, we can also prove that $x\in {\cup }_{k=1}^{\infty }{\cap }_{m=k}^{\infty }\left\{x:{\sup }_{n>m}\left|f_n(x)-f(x)\right|>1/k\right\}$.

Now we finish the proof. \end{proof}

8. Let $(X,\mathcal{A},\mu )$ be a measure space and $\left\{f_n\right\}$ a monotone sequence of measurable functions on $X$ that converges to $f$ in measure. Prove that ${\lim }_{n\to \infty }{f}_n=f$ a.e.

\begin{proof} Define $E_{m,n}=\{x\in X:|f_m(x)-f(x)|⩾\frac{1}{n}\}$, then $\{E_{m,n}{\}}_{m=1}^{\infty }$ is decreasing monotonic sequences and $\mu ({\cup }_{n=1}^{\infty }{\cap }_{m=1}^{\infty }{E}_{m,n})=0$.

Define $L:=\{x:f_n(x)\to f(x)\}$. For any $x\notin L$, since $\{f_n\}$ is monotone, $f_n(x)$ converges to $g(x)$ and $|g(x)-f(x)|>0$. It follows that there exists $M,N$ such that $x\in E_{m,n}$ for any $m>M$ and $n>N$. Therefore, $L\subseteq {\cup }_{n=1}^{\infty }{\cap }_{m=1}^{\infty }{E}_{m,n}$ and so ${\lim }_{n\to \infty }{f}_n=f$ a.e. by $\mu (L^c)=0$. \end{proof}

9. Let $(X,\mathcal{A},\mu )$ be a measure space. Let $f$ be a real-valued measurable function on a set $D\in \mathcal{A}$ with $\mu (D)<\infty$. Show that for every $ϵ>0$ there exists a set $E\in \mathcal{A}$ such that $E\subset D,\mu (D-E)<ϵ$ and $f$ is bounded on $D-E$.

\begin{proof} Since $f$ is measurable, there exists a sequence $\{f_n\}$ such that $f_n\to f$ pointwise, each $f_n$ are simple functions and $|f_n(x)|⩽|f_{n+1}(x)|⩽|f(x)|$ for all $n\in \mathbb{N}$ and all $x\in X$. By Egorov’s theorem, for any $ϵ>0$ there exists $E\subseteq D$ such that $\mu (D-E)<ϵ$ and $f_n\to f$ uniformly on $E$. Furthermore, since $f_n$ are bounded function, it yields that $f$ is bounded on $E$. \end{proof}

11. Let $f$ be a bounded real-valued measurable function on a measure space $(X,\mathcal{A},\mu )$.

• (a) Show that there exists an increasing sequence of simple functions $\left\{f_n\right\}$, such that $f_n\uparrow f$ uniformly.
• (b) Show that there exists a decreasing sequence of simple functions $\left\{g_n\right\}$, such that $g_n\downarrow f$ uniformly.

\begin{proof} a) Define $f=f^{+}-f^{-}$ where

f(x), &\mbox{if }f(x)\geqslant0\\ \\ 0,&\mbox{if }f(x)<0 \end{cases}\mbox{ and } f^-(x)=\begin{cases} 0, &\mbox{if }f(x)\geqslant0\\ \\ -f(x),&\mbox{if }f(x)<0 \end{cases}.$$ Since $f$ is bounded, we assume that $|f(x)|\leqslant M$ for all $M$. By Theorem 7.3, there exists sequences $\{f_n\}$ and $\{g_n\}$ such that $f_n,g_n$ are simple functions, $f_n\to f^+, g_n\to M-f^-$ and $|f_k(x)|\leqslant|f_{k+1}(x)|$, $|g_k(x)|\leqslant|g_{k+1}(x)|$ for all $k\in \mathbb{N}$. Since $f_n(x)\geqslant 0$ for all $x\in X$ and $g_n(x)\geqslant 0$ for all $x\in X$, it follows that $\{f_n\}$ and $\{g_n\}$ are increasing sequence. Define $h_n(x)=f_n(x)+g_n(x)-M$, then $\{h_n\}$ is an increasing sequence and $h_n\to f$. Since $f^+$ and $M-f^-$ are bounded functions, $\{f_n\}$ are $\{g_n\}$ are uniformly convergent. It follows that $\{h_n\}$ is uniformly convergent to $f$, as desired. b) Define $F(x)=-f(x)$, then $F$ is also a bounded real-valued measurable function and there exists increasing sequence $\{F_n\}$ such that $F_n\to F$ uniformly by a). Define $k_n=-F_n$, then $\{k_n\}$ is a decreasing sequence and $k_n\to f$ uniformly, as desired. `\end{proof}`