HW6

Exercise 7

12. Let $(X,\mathcal{A},\mu )$ be a measure space, $f$ an integrable function on $X$ and $A\subset X$ with $\mu (A)<\infty$. Prove that for some $x\in A$,

$$|f(x)|\le \frac{1}{\mu (A)}{\int }_A|f|d\mu .$$

\begin{proof} Otherwise, for any $x\in A$, $|f(x)|>\frac{1}{\mu (A)}{\int }_A|f|d\mu$. It follows that

$${\int }_A|f|d\mu >{\int }_A(\frac{1}{\mu (A)}{\int }_A|f|d\mu )d\mu =\mu (A)\cdot \frac{1}{\mu (A)}{\int }_A|f|d\mu ={\int }_A|f|d\mu ,$$

which is impossible. Therefore, there exists some $x\in A$ such that $|f(x)|\le \frac{1}{\mu (A)}{\int }_A|f|d\mu$. \end{proof}

15. Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space, $F\subset \mathbb{R}$ a closed set, and $f$ an integrable function on $X$ such that

$$\frac{1}{\mu (A)}{\int }_{A}fd\mu \in F\text{ for all }A\in \mathcal{A}\text{ with }\mu (A)<\infty$$

Prove that $f(x)\in F$ for a.e. $x\in X$.

\begin{proof} Since $F$ is a closed set, $F^c$ is open and $F^c$ can be written as a countable union of disjoint open intervals. Assume that $F^c={\cup }_{n=-\infty }^{\infty }(a_n,b_n)$, then $f^{-1}(F^c)={\cup }_{n=-\infty }^{\infty }{f}^{-1}((a_n,b_n))$. If $\mu (f^{-1}(F^c))\ne 0$, then there exists at least one $n_0$ such that $\mu (f^{-1}((a_{n_0},b_{n_0})))\ne 0$ and $(a_{n_0},b_{n_0})\cap F=\varnothing$.

Define $S=f^{-1}((a_{n_0},b_{n_0}))$. Since $f$ is integrable, $S\in \mathcal{A}$ and so

$$\frac{1}{\mu (S)}{\int }_{S}fd\mu \in (a_{n_0},b_{n_0}),$$

which contradicts with $(a_{n_0},b_{n_0})\cap F=\varnothing$. Therefore, $f(x)\in F$ for a.e. $x\in X$. \end{proof}

16. Let $(X,\mathcal{A},\mu )$ be a measure space and $f\in L^1(\mu )$ such that $\left|{\int }_{A}fd\mu \right|\le$ $c\mu (A)$ for all measurable $A$ with $\mu (A)<\infty$ and a constant $c$. Prove that $|f|\le c$ a.e.

\begin{proof} Define $S=f^{-1}((c,\infty ))$. If $\mu (S)\ne 0$, then

$$\frac{1}{\mu (S)}{\int }_{S}fd\mu >c,$$

which is impossible. Therefore, $\mu (S)=0$, i.e., $|f|<c$ a.e. \end{proof}

18. Give a proof of Proposition 8.7.

Proposition 8.7 Let $(X,\mathcal{A},\mu )$ be a measure space and let $f$ and $g$ be integrable, complex-valued functions defined on $X$. Let $\alpha$ and $\beta$ be complex constants. Then

$${\int }_X(\alpha f+\beta g)d\mu =\alpha {\int }_{X}fd\mu +\beta {\int }_{X}gd\mu$$

\begin{proof} Suppose $\alpha ={\alpha }_1+{\alpha }_{2}i$ and $\beta ={\beta }_1+{\beta }_{2}i$ with ${\alpha }_i,{\beta }_i\in \mathbb{R}$. Suppose $f=f_1+f_{2}i$ and $g=g_1+g_{2}i$ with $f_i,g_i$ are integrable real-valued functions. Then we have

$$\mathrm{LHS}={\int }_X({\alpha }_1{f}_1-{\alpha }_2{f}_2+{\beta }_1{g}_1-{\beta }_2{g}_2)d\mu +i{\int }_X({\alpha }_1{f}_2+{\alpha }_2{f}_1+{\beta }_1{g}_2+{\beta }_2{g}_1)d\mu$$

and

$$\mathrm{RHS}=({\alpha }_1+{\alpha }_{2}i)({\int }_X{f}_{1}d\mu +i{\int }_X{f}_{2}d\mu )+({\beta }_1+{\beta }_{2}i)({\int }_X{g}_{1}d\mu +i{\int }_X{g}_{2}d\mu ).$$

It suffices to show

$${\int }_X({\alpha }_1{f}_1-{\alpha }_2{f}_2+{\beta }_1{g}_1-{\beta }_2{g}_2)d\mu ={\alpha }_1{\int }_X{f}_{1}d\mu -{\alpha }_2{\int }_X{f}_{2}d\mu +{\beta }_1{\int }_X{g}_{1}d\mu -{\beta }_2{\int }_X{g}_{2}d\mu$$

and the proof of another equality is similar. By Theorem 8.6, we finish the proof. \end{proof}

Exercise 8

1. Suppose $\mu$ is a signed measure. Prove that $A$ is a null set with respect to $\mu$ if and only if $|\mu |(A)=0$.

\begin{proof} By Hahn decomposition, there exist disjoint $E$ and $F$ such that $E\bigsqcup F=X$, $E$ is a negative set and $F$ is a positive set. Thus $A=(A\cap E)\bigsqcup (A\cap F)$.

If $A$ is a null set, then for any $B\subseteq A$ and $B\in \mathcal{A}$ we have $\mu (B)=0$. If ${\mu }^{-}(A\cap E)\ne 0$, then $\mu (A\cap E)={\mu }^{+}(A\cap E)-{\mu }^{-}(A\cap E)\ne 0$ by ${\mu }^{+}(A\cap E)=0$, which is impossible. Hence, ${\mu }^{-}(A\cap E)=0$. Similarly, we can prove that ${\mu }^{+}(A\cap F)=0$. It deduces that ${\mu }^{+}(A)={\mu }^{+}(A\cap F)=0$ and ${\mu }^{-}(A)={\mu }^{-}(A\cap E)=0$. Therefore, $|\mu |(A)=0$.

If $|\mu |(A)=0$, then ${\mu }^{+}(A)=\mu (A\cap E)=0$ and ${\mu }^{-}(A)=\mu (A\cap F)=0$. For any subset $B\subseteq A$, we have $\mu (B)={\mu }^{+}(B)-{\mu }^{-}(B)$. Since ${\mu }^{+}$ and ${\mu }^{-}$ are measures, ${\mu }^{+}(B)⩽{\mu }^{+}(A)=0$ and ${\mu }^{-}(B)⩽{\mu }^{-}(A)=0$. Therefore, $\mu (B)=0$ for any $B\subseteq A$ and so $A$ is a null set. \end{proof}

2. Let $\mu$ be a signed measure on a measurable space $(X,\mathcal{A})$ and $f$ be a measurable function on $X$. Prove that

$$\left|{\int }_{E}fd\mu \right|\le {\int }_E|f|d|\mu |,\forall E\in \mathcal{A}.$$

\begin{proof} Recall that integration with respect to a signed measure $\mu$ is defined as

$${\int }_{X}fd\mu ={\int }_{X}fd{\mu }^{+}-{\int }_{X}fd{\mu }^{-}.$$

Since $|\mu |={\mu }^{+}+{\mu }^{-}$, we have ${\int }_E|f|d|\mu |={\int }_E|f|d{\mu }^{+}+{\int }_E|f|d{\mu }^{-}$. Then ${\int }_E|f|d{\mu }^{+}⩾\pm {\int }_{E}fd{\mu }^{+}$ and ${\int }_E|f|d{\mu }^{-}⩾\pm {\int }_{E}fd{\mu }^{-}$ yield that

$${\int }_E|f|d|\mu |={\int }_E|f|d{\mu }^{+}+{\int }_E|f|d{\mu }^{-}⩾{\int }_{E}fd{\mu }^{+}-{\int }_{E}fd{\mu }^{-}={\int }_{E}f\mu$$

and

$${\int }_E|f|d|\mu |={\int }_E|f|d{\mu }^{+}+{\int }_E|f|d{\mu }^{-}⩾-{\int }_{E}fd{\mu }^{+}+{\int }_{E}fd{\mu }^{-}=-{\int }_{E}f\mu .$$

Now we finish the proof. \end{proof}

3. Let $\mu$ be a signed measure on $(X,\mathcal{A})$. Prove that if $A\in \mathcal{A}$, then

$$|\mu |(A)=\sup \left\{\sum _{j=1}^n\left|\mu \left(B_j\right)\right|,\text{ each }{B}_j\in \mathcal{A}\text{, the }{B}_j\text{ are disjoint, }{\cup }_{j=1}^n{B}_j=A\right\}.$$

\begin{proof} By Hahn decomposition, there exist disjoint $E$ and $F$ such that $E\bigsqcup F=X$, $E$ is a negative set and $F$ is a positive set. Thus $A=(A\cap E)\bigsqcup (A\cap F)$.

For any disjoint $B_j$ with ${\bigsqcup }_{j=1}^n{B}_j=A$, there is

$$\sum _{j=1}^n|\mu (B_j)|⩽\sum _{j=1}^n|\mu |(B_j)=|\mu |(A)$$

and so

$$|\mu |(A)⩽\sup \left\{\sum _{j=1}^n\left|\mu \left(B_j\right)\right|,\text{ each }{B}_j\in \mathcal{A}\text{, the }{B}_j\text{ are disjoint, }{\cup }_{j=1}^n{B}_j=A\right\}.$$

On the other hand, for any disjoint $B_j$ with ${\bigsqcup }_{j=1}^n{B}_j=A$, define $B_{j1}=B_j\cap E$ and $B_{j2}=B_j\cap F$, then $\{B_{j1}\}\cup \{B_{j2}\}$ is a pairwise disjoint family whose union equals $A$. Note that $|\mu (B_{j1})|=|\mu (B_j\cap E)|=|{\mu }^{+}(B_j)|={\mu }^{+}(B_j)$ and $|\mu (B_{j2})|=|\mu (B_j\cap F)|=|{\mu }^{-}(B_j)|={\mu }^{-}(B_j)$, we have

$$\sum _{k=1}^2\sum _{j=1}^n|\mu (B_{jk})|=\sum _{j=1}^n|\mu |(B_j)=|\mu |(A)$$

and so $|\mu |(A)=\sup \left\{{\sum }_{j=1}^n\left|\mu \left(B_j\right)\right|,\text{ each }{B}_j\in \mathcal{A}\text{, the }{B}_j\text{ are disjoint, }{\cup }_{j=1}^n{B}_j=A\right\}$. \end{proof}

4. If $\mu ={\mu }_1-{\mu }_2$, where ${\mu }_1,{\mu }_2$ are positive measures and either ${\mu }_1$ or ${\mu }_2$ is finite, then ${\mu }_1(E)\ge {\mu }^{+}(E),{\mu }_2(E)\ge {\mu }^{-}(E)$, for all $E\in \mathcal{A}$, where $\mu ={\mu }^{+}-{\mu }^{-}$is the Jordan decomposition of $\mu$.

\begin{proof} By Hahn decomposition, there exist disjoint $M$ and $N$ such that $M\bigsqcup N=X$, $M$ is a negative set and $N$ is a positive set. Then for any $E\in \mathcal{A}$, $E=(E\cap M)\bigsqcup (E\cap N)$. Note that

$${\mu }_1(E\cap M)-{\mu }_2(E\cap M)=\mu (E\cap M)={\mu }^{+}(E\cap M)-{\mu }^{-}(E\cap M)={\mu }^{+}(E\cap M)$$

yields that ${\mu }_1(E\cap M)={\mu }_2(E\cap M)+{\mu }^{+}(E\cap M)⩾{\mu }^{+}(E\cap M)$. Also we have ${\mu }_1(E\cap N)⩾0={\mu }^{+}(E\cap N)$. Thus, ${\mu }_1(E)⩾{\mu }^{+}(E)$.

On the other hand, as ${\mu }_2(E)-{\mu }^{-}(E)={\mu }_1(E)-{\mu }^{+}(E)⩾0$, there is ${\mu }_2(E)⩾{\mu }^{-}(E)$. \end{proof}

5. Let $(X,\mathcal{A},\mu )$ be a finite signed measure space. Show that there is an $M>0$ such that $|\mu (E)|<M$ for every $E\in \mathcal{A}$.

\begin{proof} Since $\mu$ is a finite signed measure, ${\mu }^{+}(X)$ and ${\mu }^{-}(X)$ are finite. Assume that $N>0$ is a real number such that ${\mu }^{+}(X)⩽N$ and ${\mu }^{-}(X)⩽N$. For any $E\in \mathcal{A}$, we have

$$|\mu (E)|=|{\mu }^{+}(E)-{\mu }^{-}(E)|⩽{\mu }^{+}(E)+{\mu }^{-}(E)⩽{\mu }^{+}(X)+{\mu }^{-}(X)⩽N$$

and so $M=2N+1$ is what we desired. \end{proof}

7. Let $(X,\mathcal{A},\mu )$ be a signed measure space. Then, for every $E\in \mathcal{A}$; the following hold:

• (a) ${\mu }^{+}(E)=\sup \{\mu (F):F\subset E,F\in \mathcal{A}\}$;
• (b) ${\mu }^{-}(E)=\sup \{-\mu (F):F\subset E,F\in \mathcal{A}\}$.

\begin{proof} By Hahn decomposition, there exist disjoint $M$ and $N$ such that $M\bigsqcup N=X$, $M$ is a negative set and $N$ is a positive set.

(a) Note that ${\mu }^{+}(E)=\mu (E\cap M)$ and $E\cap M\subseteq E$, then we have ${\mu }^{+}(E)⩽\sup \{\mu (F):F\subset E,F\in \mathcal{A}\}$. For any $F\subseteq E$,

$$\mu (F)={\mu }^{+}(F)-{\mu }^{-}(F)⩽{\mu }^{+}(E)-{\mu }^{-}(F)⩽{\mu }^{+}(E)$$

yields that ${\mu }^{+}(E)⩾\sup \{\mu (F):F\subset E,F\in \mathcal{A}\}$. Therefore, ${\mu }^{+}(E)=\sup \{\mu (F):F\subset E,F\in \mathcal{A}\}$.

(b) Similarly, note that ${\mu }^{-}(E)=-\mu (E\cap N)$ and $E\cap N\subseteq E$, then we have ${\mu }^{-}(E)⩽\sup \{-\mu (F):F\subset E,F\in \mathcal{A}\}$. For any $F\subseteq E$,

$$\mu (F)={\mu }^{+}(F)-{\mu }^{-}(F)⩾{\mu }^{+}(F)-{\mu }^{-}(E)⩾-{\mu }^{-}(E)$$

yields that ${\mu }^{-}(E)⩾\sup \{-\mu (F):F\subset E,F\in \mathcal{A}\}$. Therefore, ${\mu }^{-}(E)=\sup \{-\mu (F):F\subset E,F\in \mathcal{A}\}$. \end{proof}

9. Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space and let $\nu$ be a measure on $(X,\mathcal{A})$ for which the conclusion of the Radon-Nikodym Theorem holds. Prove that $\nu$ is $\sigma$-finite.

\begin{proof} We aim to show $\nu (A)={\int }_{A}fd\mu$ is $\sigma$-finite, where $f$ is a finite real-valued measurable function. Since $\mu$ is $\sigma$-finite, there exists $\{E_n{\}}_{n=1}^{\infty }$ such that $E_n\uparrow X$ and $\mu (E_n)<\infty$.

As $f$ is measurable, define $E_n^{\prime }=f^{-1}([-n,n])\cap E_n$ and then $E_n^{\prime }\uparrow X$. Since $\nu (E_n^{\prime })={\int }_{E_n^{\prime }}fd\mu ⩽\mu (E_n)\cdot n<\infty$, $\nu$ is $\sigma$-finite. \end{proof}

10. Let $\mu$ and $\nu$ be measures on the measurable space $(X,\mathcal{A})$. Show that the absolute continuity condition is equivalent to $\mu (A\Delta B)=0\Rightarrow \nu (A)=\nu (B)$ for all $A,B\in \mathcal{A}$.

\begin{proof} Assume the absolute continuity condition holds, that is, if $\mu (A)=0$, then $\nu (A)=0$. If $\mu (A\Delta B)=0$, then $\nu (A\cap B^c)=\nu (A^c\cap B)=0$ and so

$$\nu (A)=\nu (A\cap B)+\nu (A\cap B^c)=\nu (A\cap B)+\nu (A^c\cap B)=\nu (B).$$

Conversely, assume $\mu (A\Delta B)=0\Rightarrow \nu (A)=\nu (B)$ for all $A,B\in \mathcal{A}$. Then if $\mu (A)=0$, then $A\Delta \varnothing =A$ and so $\nu (A)=\nu (\varnothing )=0$. \end{proof}