8 Signed Measure

Definition

Let $\mathcal{A}$ be a $\sigma$-algebra. A signed measure is a function $\mu :\mathcal{A}\to$ $(-\infty ,\infty ]$ such that

• $\mu (\varnothing )=0$;
• if $A_1,A_2,\dots$ are pairwise disjoint and all the $A_i$ are in $\mathcal{A}$, then $\mu \left({\cup }_{i=1}^{\infty }{A}_i\right)=$ ${\sum }_{i=1}^{\infty }\mu \left(A_i\right)$.

Remark.

• For any $\{A_i{\}}_{i=1}^{\infty }$, either $\{\mu (A_i)\}$ is absolutely convergent, or $\sum \mu (A_i)=\infty$.
• For example, suppose $f$ is an integrable function, then $\mu (A):={\int }_{A}f$ is a signed measure.

Proposition

The followings hold.

• Let $\mu$ be a signed measure on a measurable space $(X,\mathcal{A})$. As in the case of measure, if ${\left\{A_j\right\}}_{j=1}^{\infty }\subset \mathcal{A}$ and $A_j\uparrow {\cup }_{j=1}^{\infty }{A}_j$, then $\mu \left({\cup }_{j=1}^{\infty }{A}_j\right)={\lim }_{j\to \infty }\mu \left(A_j\right)$. Similarly, if ${\left\{A_j\right\}}_{j=1}^{\infty }\subset \mathcal{A}$ and $A_j\downarrow {\cap }_{j=1}^{\infty }{A}_j,\mu \left(A_1\right)<\infty$, then $\mu \left({\cap }_{j=1}^{\infty }{A}_j\right)=$ ${\lim }_{j\to \infty }\mu \left(A_j\right)$.
• A signed measure $\mu$ on a measurable space $(X,\mathcal{A})$ is said to be finite if $\mu (X)<$ $\infty$ and it is said to be $\sigma$-finite if there exists a sequence of sets ${\left\{A_j\right\}}_{j=1}^{\infty }\subset \mathcal{A}$ such that $X={\cup }_{j=1}^{\infty }{A}_j$ and $\mu \left(A_j\right)<\infty$ for every $j$.
• Note that if $\mu$ is a signed measure, then $\mu \left({\cup }_{i=1}^{\infty }{A}_i\right)={\lim }_{n\to \infty }\mu \left({\cup }_{i=1}^n{A}_i\right)$.

Positive/Negative Set and Hahn Decomposition Theorem

Definition

Let $\mu$ be a signed measure.

• A set $A\in \mathcal{A}$ is called a positive set for $\mu$ if $\mu (B)⩾0$ whenever $B\subset A$ and $B\in \mathcal{A}$.
• We say $A\in \mathcal{A}$ is a negative set if $\mu (B)⩽0$ whenever $B\subset A$ and $B\in \mathcal{A}$.
• A null set $A$ is one where $\mu (B)=0$ whenever $B\subset A$ and $B\in \mathcal{A}$.

Lemma

Let $\mu$ be a signed measure which takes values in $(-\infty ,\infty ]$. Let $E$ be measurable with $\mu (E)<0$. Then there exists a measurable subset $F$ of $E$ that is a negative set with $\mu (F)<0$.

\begin{proof} Assume that $E$ is not a negative set, then there exists $E_1\subseteq E$ such that $\mu (E_1)>0$ and $\mu (E_1)⩾\frac{1}{n_1}$ where $n_1$ is the smallest positive integer with this property. If $F_1=E-E_1$ is a negative set, then $F_1$ is the designed set.

Otherwise, assume that $F_1$ is not a negative set, then there exists $E_2\subseteq F_1$ such that $\mu (E_2)⩾\frac{1}{n_2}$ where $n_2$ is the smallest positive integer with this property. Assume $F_k=E-{\cup }_{i=1}^k{E}_i$ is not negative, there exists $E_{k+1}\subseteq E$ such that $\mu (E_{k+1})⩾\frac{1}{n_{k+1}}$ where $n_{k+1}$ is the smallest positive integer with this property.

Let $F=E-{\cup }_{k=1}^{\infty }{E}_k$. We claim that $F$ is a negative set with $\mu (F)<0$. Note that

$$-\infty <\mu (F)=\mu (E)-\sum _{k=1}^{\infty }\mu (E_k)⩽\mu (E)-\sum _{k=1}^{\infty }\frac{1}{n_k}<0$$

yields that ${\sum }_{k=1}^{\infty }\frac{1}{n_k}<\infty$. It follows that ${\lim }_{n\to \infty }{n}_k=\infty$.

If $n_k>1$, then for any measurable set $G\subseteq F_{k-1}$, $\mu (G)<\frac{1}{n_k-1}$. Hence, for all $H\subseteq F$, we have that $H\subseteq F_{k-1}$ for all $k\in {\mathbb{N}}_{+}$ and so $\mu (H)<\frac{1}{n_k-1}$ for all big enough $k\in {\mathbb{N}}_{+}$. Take $k\to \infty$, there is $\mu (H)⩽0$. Now we finish the proof. \end{proof}

Hahn decomposition theorem

Let $\mu$ be a signed measure taking values in $(-\infty ,\infty ]$.

• There exist disjoint measurable sets $E$ and $F$ in $\mathcal{A}$ whose union is $X$ and such that $E$ is a negative set and $F$ is a positive set.
• If $E_0$ and $F_0$ are another such pair, then $E\Delta E_0=F\Delta F_0$ is a null set with respect to $\mu$.
• If $\mu$ is not a positive measure, then $\mu (E)<0$. If $-\mu$ is not a positive measure, then $\mu (F)>0$.

Remark. We write $A\Delta B$ for $(A-B)\cup (B-A)$.

\begin{proof} i) Assume that there exists measurable set with negative measure. Let $\mathcal{F}=\{A\in \mathcal{A}:A\text{mbox}isanegativeset\}$. Let $L={\inf }_{A\in \mathcal{A}}\mu (A)<0$. Let $\{A_i\}\subseteq \mathcal{F}$ with $\mu (A_i)\to L$. Let $B_n=A_n-{\cup }_{i=1}^{n-1}{A}_i$. Then $B_i$ is a negative set and $B_i\cap B_j=\varnothing$. Define $A={\cup }_{i=1}^{\infty }{A}_i={\cup }_{i=1}^{\infty }{B}_i$. For any $C\subseteq A$, $\mu (C)={\sum }_{n=1}^{\infty }\mu (B_n\cap C)<0$ and so $A\in \mathcal{F}$. Note that

$$L⩽\mu (A)=\mu (A\setminus A_n)+\mu (A_n)⩽\mu (A_n)\to L\text{mbox}asn\to \infty .$$

Therefore, $\mu (A)=L$.

Let $E=A$ and $F=E^c$. Then $E,F$ is a partition of $X$. Now we claim that $F$ is a positive set. Otherwise, there is $G\in \mathcal{A}$ such that $\mu (G)<0$. By ^6e1e47, we can assume $G$ is a negative set and then $\mu (E\cup G)<\mu (E)$, which is a contradiction.

ii) If $E_0,F_0\in \mathcal{A}$ with $E_0,F_0$ another pair, then $E-E_0=F_0-F$ and $E_0-E=F-F_0$. Thus $E\Delta E_0=F\Delta F_0$. Let $A\subseteq E-E_0=F_0-F$. It follows that $\mu (A)⩾0$, $\mu (A)⩽0$ and so $\mu (A)=0$. Therefore, $E\cap E_0$ is null.

iii) If $\mu$ is not a positive measure and $\mu (E)=0$, for any $A\in \mathcal{A}$, $\mu (A)=\mu (A\cap E)+\mu (A\cap F)=\mu (A\cap F)⩾0$, which contradicts with $\mu$ not positive. \end{proof}

Jordan Decomposition Theorem

Definition

We say two measures $\mu$ and $\nu$ are mutually singular if there exist two disjoint sets $E$ and $F$ in $\mathcal{A}$ whose union is $X$ with $\mu (E)=\nu (F)=0$. This is often written as $\mu \perp \nu$.

Jordan Decomposition Theorem

If $\mu$ is a signed measure on a measurable space $(X,\mathcal{A})$, there exist positive measures ${\mu }^{+}$and ${\mu }^{-}$such that $\mu ={\mu }^{+}-{\mu }^{-}$and ${\mu }^{+}$and ${\mu }^{-}$are mutually singular. This decomposition is unique.

\begin{proof} i) Let $E,F\in \mathcal{A}$ be a partition of $X$ where $E$ is negative of $\mu$ and $F$ is positive of $\mu$. Define ${\mu }^{+},{\mu }^{-}$ by ${\mu }^{+}(A)=\mu (A\cap F)$ and ${\mu }^{-}(A)=\mu (A\cap E)$. Then ${\mu }^{+}\perp {\mu }^{-}$.

ii) Assume that $\mu ={\mu }^{+}-{\mu }^{-}={\nu }^{+}-{\nu }^{-}$ with ${\mu }^{+}\perp {\mu }^{-}$ and ${\nu }^{+}\perp {\nu }^{-}$, and assume that $E_0,F_0$ with ${\nu }^{+}(E_0)={\nu }^{-}(F_0)=0$. For any $A\in \mathcal{A}$ with $A\subseteq E_0$, note that

$$\mu (A)={\nu }^{+}(A)-{\nu }^{-}(A)=0-{\nu }^{-}(A)⩽0.$$

If $A\in \mathcal{A}$ with $A\subseteq F_0$, then $\mu (A)⩾0$. Hence $E_0,F_0$ give another Hahn-decomposition of $\mu$ and so $E\Delta E_0=F\Delta F_0$ is null. For any $A\in \mathcal{A}$, ${\nu }^{+}(A)={\nu }^{+}(A\cap F_0)+{\nu }^{+}(A\cap E_0)={\nu }^{+}(A\cap F_0)$ and so $\mu (A\cap F_0)={\nu }^{+}(A)$. Further, ${\mu }^{+}(A)=\mu (A\cap F_0)$ and it follows that ${\nu }^{+}={\mu }^{+}$ for any $A\in \mathcal{A}$. \end{proof}

Remark. Since $\mu$ takes values in $(-\infty ,\infty ]$, from the construction of ${\mu }^{\pm }$in the proof of Theorem 9.7, we know that ${\mu }^{-}$ is a finite measure. Thus if $\mu$ is a finite (respectively, $\sigma$-finite), then the same is true for ${\mu }^{\pm }$.

Absolutely Continuous

Definition

A measure $\nu$ is said to be absolutely continuous with respect to a measure $\mu$ if $\mu (A)=0\Rightarrow \nu (A)=0$. In this case, we write $\nu \ll \mu$.

Lemma

Let $\nu$ be a finite measure. Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if for all $ϵ>0$ there exists $\delta >0$ such that $\mu (A)<\delta$ implies $\nu (A)<ϵ$.

\begin{proof} One direction is easy. Otherwise, assume that there exists ${ϵ}_0>0$ such that for and $\delta >0$, there is $A$ with $\mu (A)<\delta$ and $\nu (A)⩾{ϵ}_0$. By taking $\delta =1/2^k$, there is $A_k$ such that $\mu (A_k)<1/2^k$ and $\nu (A_k)⩾ϵ$. It follows that $E={\cap }_{n=1}^{\infty }{\cup }_{k=n}^{\infty }{A}_k$ satisfies $\mu (E)=0$ and $\nu (E)⩾{ϵ}_0$, which is a contradiction. \end{proof}

Radon-Nikodym Theorem

Proposition

Let $\mu$ and $\nu$ be finite positive measures on a measurable space $(X,\mathcal{A})$. Then either $\mu \perp \nu$ or else there exist $ϵ>0$ and $G\in \mathcal{A}$ such that $\mu (G)>0$ and $G$ is a positive set for $\nu -ϵ\mu$.

\begin{proof} Consider $\nu -\frac{1}{n}\mu$. Let $E_n,F_n\in \mathcal{A}$ such that $E_n\cap F_n=\varnothing$, $E_n\cup F_n=X$, where $E_n$ is negative for $\nu -\frac{1}{n}\mu$ and $F_n$ is positive for $\nu -\frac{1}{n}\mu$. Let $E={\cap }_{n=1}^{\infty }{E}_n$, and let $F={\cup }_{n=1}^{\infty }{F}_n$. Then $E^c=F$ and $(\nu -\frac{1}{n}\mu )(E_n)⩽0$ for all $n$. It follows that $\nu (E_n)⩽\frac{1}{n}\mu (E_n)⩽\frac{1}{n}\mu (X)$ and so $\nu (E)⩽\nu (E_n)\to 0$ as $n\to \infty$. Therefore, $\nu (E)=0$.

• if $\mu (E^c)=\mu (F)=0$, then $\nu \perp \mu$.
• if $\mu (F)>0$, then there exists $n$ with $\mu (F_n)>0$. Let $ϵ=\frac{1}{n}$, and let $G=F_n$. Then $\mu (G)>0$ and so $G$ is positive of $\nu -ϵ\mu$.

Now we finish the proof. \end{proof}

Radon-Nikodym theorem

Suppose $\mu$ and $\nu$ are $\sigma$-finite positive measures on a measurable space $(X,\mathcal{A})$ such that $\nu$ is absolutely continuous with respect to $\mu$. Then there exists a $\mu$-measurable non-negative function $f$ such that

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

for all $A\in \mathcal{A}$. Moreover, if $g$ is another such function, then $f=g$ almost everywhere with respect to $\mu$.

\begin{proof} i) Uniqueness. If $g,f⩾0$ satisfy the same condition, then ${\int }_A(f-g)d\mu =0$ for all $A\in \mathcal{A}$. Thus, $f=g$ almost everywhere with respect to $\mu$.

ii) Let $\mu ,\nu$ be finite. Let $\mathcal{J}=\{f:f⩾0\text{mbox}measurable,\nu (A)⩾{\int }_{A}fd\mu ,\forall A\in \mathcal{A}\}$. Since $0\in \mathcal{J}$, $\mathcal{J}\ne \varnothing$. Let $L={\sup }_{g\in \mathcal{J}}{\int }_{X}gd\mu$. Let $\{g_n\}\subseteq \mathcal{J}$. We need only to consider $h_2$. Since $h_2=\max g_1,g_2$, for any $A\in \mathcal{A}$, consider $A_1=\{x\in A:g_1(x)⩾g_2(x)\}$ and $A_2=\{x\in A:g_1(x)<g_2(x)\}$. Since

$${\int }_A{h}_{2}d\mu ={\int }_{A_1}{g}_{1}d\mu +{\int }_{A_2}{g}_{2}d\mu ⩽\nu (A_1)+\nu (A_2)=\nu (A),$$

we have that $h_2\in \mathcal{J}$. Note that $h_n\uparrow h$, so we have that ${\int }_X{g}_{n}d\mu ⩽{\int }_X{h}_{n}d\mu ⩽{\int }_{X}hd\mu$. Take $n\to \infty$, then $L⩽{\int }_{X}hd\mu ⩽L$. Therefore, ${\int }_{X}hd\mu =L$, i.e. there exists $f\in \mathcal{J}$ such that ${\int }_{X}fd\mu =L={\sup }_{g\in \mathcal{J}}{\int }_{x}gd\mu$.

Consider $\lambda (A)=\nu (A)-{\int }_{X}fd\mu$, then $\lambda$ is a positive measure on $(X,\mathcal{A})$. By ^37ff4c, one of the following holds:

• $\lambda \perp \mu$
• there exists $G\in \mathcal{A}$, $ϵ>0$ such that $\mu (G)>0$ and $G$ is a positive set of $\lambda -ϵ\mu$.

If the second case is true, then for any $A\in \mathcal{A}$, there is

$$\nu (A)-{\int }_{A}fd\mu =\lambda (A)⩾\lambda (A\cap G)⩾ϵ\mu (G\cap A)={\int }_Aϵ{\chi }_{G}d\mu$$

and $\nu (A)⩾{\int }_A(f+ϵ{\chi }_G)d\mu$. Therefore, $f+ϵ{\chi }_G\in \mathcal{J}$. However, ${\int }_X(f+ϵ{\chi }_G)d\mu =L+ϵ\mu (G)>L$, which is a contradiction.

It deduces that $\lambda \perp \mu$. Therefore, there exists $H,H^c\in \mathcal{A}$ such that $\lambda (H)=\mu (H^c)=0$ and $\lambda (H^c)=\nu (H^c)-{\int }_{H^c}fd\mu =0-0=0$. Therefore, $\nu (A)={\int }_{A}fd\mu$ for all $A\in \mathcal{A}$.

iii) Let $\mu ,\nu$ be $\sigma$-finite. There exists measurable $X_n$ with $X_n\uparrow X$ such that $\mu (X_n)<\infty$ and $\nu (X_n)<\infty$. By ii), there exists $f_n$ such that $\nu (A)={\int }_A{f}_{n}d\mu$ for any $A\subseteq X_n$ and $A\in \mathcal{A}$. Note that $f_{n+1}=f_n$ on $X_n$ almost everywhere. Define $f:X\to [0,\infty ]$ by $f{|}_{X_n}=f_n$. It is easy to show that $f$ is measurable. Furthermore, note that

$$\nu (A)=\underset{n\to \infty }{\lim }\nu (A\cap X_n)=\underset{n\to \infty }{\lim }{\int }_{A\cap X_n}{f}_n=\underset{n\to \infty }{\lim }{\int }_{A}f{\chi }_{X_n}d\mu ={\int }_{A}fd\mu .$$

Now we finish the proof. \end{proof}

Remark. The hypothesis that $(X,\mathcal{A},\mu )$ is $\sigma$-finite cannot be removed, see here.

Definition

Let $(X,\mathcal{A})$ be a measurable space and $\mu$ and $\nu$ be signed measures on $X$. We say that $\nu$ is absolutely continuous with respect to $\mu$ if $|\mu |(A)=$ $0\Rightarrow \nu (A)=0$. In this case, we also write $\nu \ll \mu$.

Radon-Nikodym

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space and $\nu$ be a $\sigma$-finite signed measure defined on $\mathcal{A}$ such that $\nu \ll \mu$. Then, there exists a measurable function $f$ defined on $X$ such that

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

for every $A\in \mathcal{A}$. The function $f$ is unique in the sense that if $g$ is any real-valued measurable function on $X$ with $\nu (A)={\int }_{A}gd\mu$ for all $A\in \mathcal{A}$, then $f=g$ a.e. respect to $\mu$.

\begin{proof} Let $\nu ={\nu }^{+}-{\nu }^{-}$ be the Jordan decomposition of $\nu$, i.e. there exists $E,F\in \mathcal{A}$ such that ${\nu }^{+}(E)={\nu }^{-}(F)=0$, $E\cap F=\varnothing$ and $E\cup F=X$. Note that ${\nu }^{-}$ is finite and ${\nu }^{+}$ is $\sigma$-finite. If $|\mu |(A)=0$, then $|\mu |(A\cap F)=0$, $|\mu |(A\cap E)=0$ and $\nu (A\cap F)=\nu (A\cap E)=0$. It follows that ${\nu }^{+}\ll \mu$ and ${\nu }^{-}\ll \mu$. By ^bdb1bb, there exists $f_1,f_2$ such that ${\nu }^{+}(A)={\int }_A{f}_{1}d\mu$ and ${\nu }^{-}(A)={\int }_A{f}_{2}d\mu$ for all $A\in \mathcal{A}$. Let $f=f_1-f_2$.

Consider the decomposition $f=f^{+}-f^{-}$, and we claim that $f^{+}⩽f_1$ and $f^{-}⩽f_2$. If $f(x)>0$, then $f^{+}(x)=f(x)⩽f_1(x)$. If $f(x)⩽0$, then $0=f^{+}(x)⩽f_1(x)$. Thus $f^{+}(x)⩽f_1(x)$. If $f(x)⩽0$, then $f^{-}(x)=-f(x)=f_2(x)-f_1(x)⩽f_2(x)$. Thus $f^{-}(x)⩽f_2(x)$ for all $x\in X$. Now we prove the claim. With this claim and ${\nu }^{-}(X)<\infty$, we deduce that${\int }_X{f}^{-}d\mu ⩽{\int }_X{f}_{2}d\mu <\infty$ and so ${\int }_{X}fd\mu$ is well-defined.

Note that ${\int }_A{f}^{+}+f_2={\int }_A{f}_1+f^{-}$, it follows that

$${\int }_{A}fd\mu ={\int }_A{f}_{1}d\mu -{\int }_A{f}_{2}d\mu ={\nu }^{+}(A)-{\nu }^{-}(A)=\nu (A)$$

and so $f$ is the function we desire. \end{proof}