HW5

1. Suppose $f$ is a non-negative integrable function on a measure space $(X,\mathcal{A},\mu )$. Prove that ${\lim }_{t\to \infty }t\mu (\{x:f(x)\ge t\})=0$.

\begin{proof} Define $E_t=\{x:f(x)⩾t\}$. Note that $f{\chi }_{E_t}⩽f$. Since $f$ is integrable, by DCT we have

$$\underset{t\to \infty }{\lim }\int f{\chi }_{E_t}\to \int \underset{t\to \infty }{\lim }f{\chi }_{E_t}=0.$$

For any $x\in X$ such that $f(x)<\infty$, there exists $t\in \mathbb{N}$ with $t>f(x)$ and $f(x){\chi }_t(x)=0$. It follows that

$$\left\{x:f(x)<\infty \right\}\subseteq \left\{x\in X\mid \underset{t\to \infty }{\lim }f(x){\chi }_{E_t}(x)=0\right\}=M$$

and the set $M$ has a negligible complement. Therefore, $t\mu (\{x:f(x)⩾t\})=t\mu (E_t)⩽{\int }_{E_t}fd\mu =\int f{\chi }_{E_t}\to 0$ as $t\to \infty$. \end{proof}

2. Let $X$ be a set and $\mathcal{A}=P(X)$, the collection of all subsets of X . Pick $x\in X$ and let ${\delta }_x$ be the Dirac measure at $x$. Prove that if $f:X\to \mathbb{R}$, then

$${\int }_{X}fd{\delta }_x=f(x).$$

\begin{proof} Let $g$ denote the constant function being equal to $f(x)$. Notice that

$${\delta }_x(\{y:g(y)=f(y)\})=1$$

as $x\in \{y:g(y)=f(y)\}$. Thus $f=g$ almost everywhere respect to Dirac measure and it follows that

$$\int f\mathrm{d}{\delta }_x=\int g\mathrm{d}{\delta }_x=f(x)\int 1\mathrm{d}{\delta }_x=f(x)\cdot {\delta }_x(X)=f(x).$$

Now we finish the proof. \end{proof}

3. Let $(X,\mathcal{A},\mu )$ be a measure space and let $f:X\to [0,\infty ]$ be a measurable function. Then the function

$${\mu }_f:\mathcal{A}\to [0,\infty ]$$

defined by

$${\mu }_f(A)={\int }_{A}fd\mu$$

is a measure, and

$${\int }_{E}gd{\mu }_f={\int }_{E}gfd\mu$$

for every measurable function $g:X\to [0,\infty ]$ and every $E\in \mathcal{A}$.

\begin{proof} i) First we verify ${\mu }_f$ is a measure. Since $\mu (\varnothing )=0$, $\mu (A)={\int }_{A}fd\mu ⩾0$ for all $A\in \mathcal{A}$ and ${\mu }_f({\bigsqcup }_{i=1}^{\infty }{A}_i)={\int }_{{\bigsqcup }_{i=1}^{\infty }{A}_i}fd\mu ={\sum }_{i=1}^{\infty }{\int }_{A_i}fd\mu ={\sum }_{i=1}^{\infty }{\mu }_f(A_i)$, ${\mu }_f$ is a measure.

ii) Since $g$ is a measurable function, there exists simple functions $\{s_n\}$ such that $s_n\uparrow g$ and so $s_{n}f\uparrow gf$. For any simple function $s$, assume that $s={\sum }_{k=1}^m{a}_k{\chi }_{F_k}$ and so

$${\int }_{E}sd{\mu }_f=\sum _{k=1}^m{a}_k{\int }_{E\cap F_k}d{\mu }_f=\sum _{k=1}^m{a}_k{\int }_{F_k\cap E}fd\mu ={\int }_E\sum _{k=1}^m{a}_k{\chi }_{F_k}fd\mu =\int sfd\mu .$$

Then we have ${\int }_E{s}_{n}d{\mu }_f={\int }_E{s}_{n}fd\mu$ for all $E\in \mathcal{A}$ and $s_n\in \{s_n\}$. Therefore, by MCT we have ${\int }_{E}gd{\mu }_f={\int }_{E}gfd\mu$ for all $E\in \mathcal{A}$. \end{proof}

4. If ${\int }_{X}fd\mu >0$, where $f$ is a measurable extended real-valued function on $X$, show that $\{x\in X:f(x)>0\}$ has positive measure.

\begin{proof} Define $A:=\{x:f(x)>0\}$, then $A^c=\{x:f(x)⩽0\}$. If $\mu (A)=0$, then there is

$$0<{\int }_{X}fd\mu ={\int }_{A}fd\mu +{\int }_{A^c}fd\mu ={\int }_{X}f{\chi }_{A}d\mu +{\int }_{A^c}fd\mu ={\int }_{A^c}fd\mu ⩽0,$$

which is a contradiction. Therefore, $A$ has positive measure. \end{proof}

5. Let $\left\{f_n\right\}$ and $\left\{g_n\right\}$ be sequences of measurable functions on $X$ such that $\left|f_n\right|\le g_n$ for every $n$. Let $f$ and $g$ be measurable functions such that ${\lim }_n{f}_n=f$ a.e. and ${\lim }_n{g}_n=g$ a.e. If ${\lim }_n{\int }_X{g}_n={\int }_{X}g<\infty$, show that

$$\underset{n}{\lim }{\int }_X{f}_n={\int }_{X}f.$$

\begin{proof} By Theorem 8.11, the statement holds. \end{proof}

6. Let $\left\{f_n\right\}$ be sequences of integrable functions on $X$ such that $f_n\to f$ a.e., where $f$ is also integrable. Show that ${\lim }_n{\int }_X\left|f_n-f\right|d\mu =0$ iff ${\lim }_n{\int }_X\left|f_n\right|d\mu =$ ${\int }_X|f|d\mu$.

\begin{proof} Assume that ${\lim }_n{\int }_X|f_n-f|d\mu =0$. To show ${\lim }_n{\int }_X\left|f_n\right|d\mu =$ ${\int }_X|f|d\mu$, it suffices to show ${\lim }_n{\int }_X||f_n|-f|d\mu =0$. Note that $0⩽||f_n|-f|⩽|f_n-f|\to 0$ as $n\to \infty$ and ${\lim }_X|f_n-f|d\mu =0$, then by GDCT,

$$\underset{n}{\lim }{\int }_X||f_n|-f|d\mu ={\int }_X\underset{n}{\lim }||f_n|-f|d\mu =0.$$

Assume that ${\lim }_n{\int }_X\left|f_n\right|d\mu =$ ${\int }_X|f|d\mu$, then ${\lim }_n{\int }_X(|f_n|-|f|)d\mu =0$. Since $|f_n-f|⩽|f_n|+|f|$ and $|f_n|+|f|\to 2|f|$ a.e., by GDCT we have

$$\underset{n}{\lim }{\int }_X|f_n-f|d\mu ={\int }_X\underset{n}{\lim }|f_n-f|d\mu =0.$$

Now we finish the proof. \end{proof}

8. Suppose $\mu (X)<\infty ,\left\{f_n\right\}$ is a sequence of bounded complex measurable functions on $X$, and $f_n\to f$ uniformly on $X$. Prove that

$$\underset{n\to \infty }{\lim }{\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu .$$

\begin{proof} Assume that $f_n=g_n+i{h}_n$ where $g_n,h_n$ are bounded real measurable functions. Also assume that $f=g+ih$ with real functions $g,h$. Then $g_n\to g$ and $h_n\to h$ uniformly on $X$. To show ${\lim }_{n\to \infty }{\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu$, it is enough to show ${\lim }_{n\to \infty }{\int }_X{g}_{n}d\mu ={\int }_{X}gd\mu$ and ${\lim }_{n\to \infty }{\int }_X{h}_{n}d\mu ={\int }_{X}hd\mu$. Here we only prove the former one.

Since $g_n\to g$ uniformly on $X$, for any $ϵ>0$ there exists $N$ such that $|g_n-g|<ϵ/\mu (X)$ for all $x\in X$ and so

$${\int }_X{g}_{n}d\mu -{\int }_{X}gd\mu ⩽{\int }_X|g_n-g|d\mu <ϵ/\mu (X)\cdot \mu (X)=ϵ.$$

Therefore, ${\lim }_{n\to \infty }{\int }_X{g}_{n}d\mu ={\int }_{X}gd\mu$ and now we finish the proof. \end{proof}

9. Let $(X,\mathcal{A},\mu )$ be a measure space. Show that if $f\in L^1(\mu )$, then

$$\underset{n\to \infty }{\lim }{\int }_{\{|f|>n\}}|f|d\mu =0.$$

\begin{proof} Since $f\in L^1(\mu )$, we have ${\lim }_n{\int }_X|f|<\infty$. Then by Chebyshev’s inequality, we have

$$\mu (\{x:|f(x)|>n\})⩽\frac{1}{n}{\int }_X|f|d\mu .$$

Define $E_n=\{x:|f(x)|⩽n\}$, then $\mu (E_n^c)⩽\frac{1}{n}{\int }_X|f|d\mu \to 0$ as $n\to \infty$. Let $f_n=|f|{\chi }_{E_n}$, then ${\lim }_n{f}_n=|f|$ almost everywhere. By MCT we have

$$\underset{n}{\lim }{\int }_X{f}_n{\chi }_{E_n}={\int }_X|f|<\infty$$

and it yields that ${\int }_{\{|f|>n\}}|f|d\mu ={\int }_{E^c}|f|={\int }_X(|f|-|f|{\chi }_{E_n})d\mu \to 0$ as $n\to \infty$. \end{proof}

10. Let $(X,\mathcal{A},\mu )$ be a finite measure space and $f$ a measurable function on $X$. For an integer $n$, let $A_n=\{|f|>n\}$ and $B_n=\{n\le |f|<n+1\}$. Prove that the following statements are equivalent: a) $f\in L^1(\mu )$; b) ${\sum }_{n}n\mu \left(B_n\right)<\infty$; ${\sum }_n\mu \left(A_n\right)<\infty$.

\begin{proof} a)→b). Assume that $f\in L^1(\mu )$, then we have ${\int }_X|f|d\mu <\infty$. Note that

$${\int }_X|f|d\mu =\sum _{n=0}^{\infty }{\int }_{B_n}|f|d\mu ⩾\sum _{n=0}^{\infty }n\mu (B_n)$$

yields that ${\sum }_{n}n\mu \left(B_n\right)<\infty$. For each $n$, we have $\mu (A_n)={\sum }_{k=n}^{\infty }\mu (B_k)$ and ${\sum }_n\mu (A_n)={\sum }_{n}n\mu (B_n)<\infty$. In fact, ${\sum }_{n}n\mu \left(B_n\right)<\infty$ and ${\sum }_n\mu \left(A_n\right)<\infty$ are equivalent.

b)→a). Assume that ${\sum }_{n}n\mu \left(B_n\right)<\infty$ holds, then

$${\int }_X|f|d\mu =\sum _{n=0}^{\infty }{\int }_{B_n}|f|d\mu ⩽\sum _{n=0}^{\infty }(n+1)\mu (B_n)⩽2\sum _{n=0}^{\infty }n\mu (B_n)+\mu (B_0)⩽2\sum _{n=0}^{\infty }n\mu (B_n)+\mu (X)<\infty$$

and so $f\in L^1(X)$. \end{proof}