16.1 Signed and Complex Measures
Definition 16.1.1 (Signed Measure). Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$, then $\mu$ is a signed measure if
$\mu(\emptyset) = 0$.
For any $\seq{E_n}\subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu(E_{n})$ where the sum converges absolutely.
By Riemann’s Rearrangement Theorem, (M2) implies that $\mu$ can only take at most one value in $\bracs{-\infty, \infty}$.
Definition 16.1.2 (Complex Measure). Let $(X, \cm)$ be a measurable space and $\mu: \cm \to \complex$, then $\mu$ is a complex measure if
$\mu(\emptyset) = 0$.
For any $\seq{E_n}\subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu(E_{n})$ where the sum converges absolutely.
Definition 16.1.3 (Positive/Negative/Null Sets). Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, and $A \in \cm$, then $A$ is...
positive if $\mu(B) \ge 0$ for all $B \in \cm$ with $B \subset A$.
negative if $\mu(B) \le 0$ for all $B \in \cm$ with $B \subset A$.
null if $\mu(B) = 0$ for all $B \in \cm$ with $B \subset A$.
Proposition 16.1.4. Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then:
For any $\seq{E_n}\subset \cm$ and $E \in \cm$ with $E_{n} \upto E$, $\mu\paren{\bigcup_{n \in \natp}E_n}= \limv{n}\mu(E_{n})$.
For any $\seq{E_n}\subset \cm$ and $E \in \cm$ with $E_{n} \downto E$ and $|\mu(E_{1})| < \infty$, $\mu\paren{\bigcap_{n \in \natp}E_n}= \limv{n}\mu(E_{n})$.
Proof. (1): For each $N \in \natp$, let $F_{N} = E_{N} \setminus \bigcup_{n = 1}^{N-1}E_{n}$, then $E_{N} = \bigsqcup_{n = 1}^{N} F_{n}$. By (M2),
(2): By (1),
Lemma 16.1.5. Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then
For any $A \in \cm$ positive and $B \in \cm$ with $B \subset A$, $B$ is positive.
For any $\seq{A_n}\subset \cm$ positive, $\bigcup_{n \in \natp}A_{n}$ is positive.
Proof. (1): For any $C \in \cm$ with $C \subset B$, $C \subset A$, so $\mu(C) \ge 0$. Hence $B$ is positive.
(2): For each $N \in \natp$, let $B_{N} = A_{N} \setminus \sum_{n = 1}^{N-1}A_{n}$. By (1), $B_{N}$ is positive with $\bigcup_{n \in \natp}A_{n} = \bigsqcup_{n \in \natp}B_{n}$. For any $C \subset \bigcup_{n \in \natp}B_{n}$,
Theorem 16.1.6 (Hahn Decomposition). Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then there exists $P, N \in \cm$ such that:
$P$ is positive.
$N$ is negative.
$X = P \sqcup N$.
For any $P', N' \in \cm$ satisfying (1)-(3), $P \Delta P'$ and $N \Delta N'$ are null.
The disjoint union $X = P \sqcup N$ is the Hahn decomposition of $\mu$.
Proof [Theorem 3.3, Fol99]. By flipping the sign of $\mu$, assume without loss of generality that $\mu < \infty$.
(1): Let $M = \sup\bracs{\mu(P)|P \in \cm \text{ positive}}$, then there exists positive sets $\seq{P_n}\subset \cm$ such that $\mu(P_{n}) \upto M$. Let $P = \bigcup_{n \in \natp}P_{n}$, then $P$ is positive by Lemma 16.1.5, and $\sup_{n \in \natp}\mu(P_{n}) \le \mu(P) \le M$.
(3): Let $N = X \setminus P$, then $P$ is positive and $X = P \sqcup N$.
(2): Suppose for contradiction that $N$ is not negative, then
Let $A_{1} \in \cm$ such that $A_{1} \subset N$ and $\mu(A_{1}) > 0$. Since $A_{1}\cap P = \emptyset$ and $\mu(P) = M$, $A_{1}$ cannot be positive, so $M_{1} = m(A_{1}) - \mu(A_{1}) > 0$. Let $n \in \natp$ and suppose inductively that $\bracs{A_n}_{1}^{n}$ has been constructed such that
For each $1 \le k \le n$, $M_{k} = m(A_{k}) - \mu(A_{k}) > 0$.
For each $1 \le k \le n - 1$, $A_{k+1}\subset A_{k}$ with $\mu(A_{k+1}) - \mu(A_{k}) > M_{k}/2$.
Let $A_{n} \in \cm$ with $A_{n+1}\subset A_{n}$ such that $\mu(A_{n+1}) - \mu(A_{n}) > M_{n}/2$. Since $A_{n+1}\cap P = \emptyset$ and $\mu(P) = M$, $A_{n+1}$ cannot be positive, so
Let $A = \bigcap_{n \in \natp}A_{n}$, then since $0 < \mu(A_{2}) < \infty$, $\mu(A) = \limv{n}\mu(A_{n})$ by Proposition 16.1.4, so
Since $A \cap P = \emptyset$ and $\mu(P) = M$, $A$ cannot be positive. Thus there exists $B \in \cm$ with $B \subset A$ with $\mu(B) - \mu(A) > 0$. As $M_{n} \to 0$ as $n \to \infty$, there exists $n \in \natp$ such that
As $\mu(B) - \mu(A_{n}) > \mu(B) - \mu(A)$, this contradicts the definition of $m(A_{n})$.
Therefore $N$ must be negative, and $X = P \sqcup N$ is the desired decomposition.
(U): Since $X = P \sqcup N = P' \sqcup N'$,
is a union of two null sets, which is null. Likewise,
is also a null set.$\square$
Corollary 16.1.7. Let $(X, \cm)$ be a measurable space, then
For any signed measure $\mu: \cm \to (-\infty, \infty)$, $\mu$ is bounded.
For any complex measure $\mu: \cm \to \complex$, $\mu$ is bounded.
Proof. (1): Let $X = P \sqcup N$ be a Hahn decomposition of $\mu$, then for any $E \in \cm$, $\mu(E) \in [\mu(N), \mu(P)]$.
(2): Since $\text{Re}(\mu)$ and $\text{Im}(\mu)$ are both signed measures taking values in $(-\infty, \infty)$, they are bounded by (1). Thus $\mu$ is also bounded.$\square$
Theorem 16.1.8 (Jordan Decomposition). Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure on $X$, then there exists positive measures $\mu^{+}, \mu^{-}: \cm \to [0, \infty]$ such that
$\mu^{+} \perp \mu^{-}$.
$\mu = \mu^{+} - \mu^{-}$.
For any other pair $(\nu^{+}, \nu^{-})$ satisfying (1) and (2), $\mu^{+} = \nu^{+}$ and $\mu^{-} = \nu^{-}$.
Proof. (1), (2): Let $X = P \sqcup N$ be a Hahn decomposition of $\mu$. For each $E \in \cm$, let
then $(\mu^{+}, \mu^{-})$ satisfies (1) and (2).
(U): Let $X = P' \sqcup N'$ such that $P'$ is $\nu^{-}$-null and $N'$ is $\nu^{+}$-null, then $X = P' \sqcup N'$ is a Hahn decomposition of $\mu$. Since $P' \Delta P = 0$ and $N' \Delta N = 0$, for any $E \in \cm$,
Likewise,