\section{Signed and Complex Measures}
\label{section:signed-complex-measure}

\begin{definition}[Signed Measure]
\label{definition:signed-measure}
    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$, then $\mu$ is a \textbf{signed measure} if
    \begin{enumerate}
        \item[(M1)] $\mu(\emptyset) = 0$.
        \item[(M2)] 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. 
    \end\{enumerate\}

    By \hyperref[Riemann's Rearrangement Theorem]{theorem:riemann-rearrangement}, (M2) implies that $\mu$ can only take at most one value in $\bracs{-\infty, \infty}$. 
\end{definition}


\begin{definition}[Complex Measure]
\label{definition:complex-measure}
    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to \complex$, then $\mu$ is a \textbf{complex measure} if
    \begin{enumerate}
        \item[(M1)] $\mu(\emptyset) = 0$.
        \item[(M2)] 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. 
    \end{enumerate}
\end{definition}


\begin{definition}[Positive/Negative/Null Sets]
\label{definition:positive-negative-sets}
    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, and $A \in \cm$, then $A$ is...
    \begin{enumerate}
        \item \textbf{positive} if $\mu(B) \ge 0$ for all $B \in \cm$ with $B \subset A$.
        \item \textbf{negative} if $\mu(B) \le 0$ for all $B \in \cm$ with $B \subset A$.
        \item \textbf{null} if $\mu(B) = 0$ for all $B \in \cm$ with $B \subset A$.
    \end{enumerate}
\end{definition}


\begin{proposition}
\label{proposition:signed-measure-properties}
    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then:
    \begin{enumerate}
        \item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \upto E$, $\mu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
        \item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \downto E$ and $|\mu(E_1)| < \infty$, $\mu\paren{\bigcap_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
    \end{enumerate}
\end{proposition}
\begin{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), 
    \[
    \mu(E) = \limv{N}\sum_{n = 1}^N \mu(F_N) = \limv{N}\mu(E_N)
    \]

    (2): By (1), 
    \[
    \mu(E) = \mu(E_1) - \mu(E_1 \setminus E) = \limv{n}[\mu(E_1) - \mu(E_1 \setminus E_n)] = \limv{n}\mu(E_n)
    \]
\end{proof}


\begin{lemma}
\label{lemma:positive-sets-properties}
    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then
    \begin{enumerate}
        \item For any $A \in \cm$ positive and $B \in \cm$ with $B \subset A$, $B$ is positive. 
        \item For any $\seq{A_n} \subset \cm$ positive, $\bigcup_{n \in \natp}A_n$ is positive. 
    \end{enumerate}
\end{lemma}
\begin{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 $C \subset \bigcup_{n \in \natp}B_n$, 
    \[
        \mu(C) = \sum_{n \in \natp}\mu(C \cap B_n) \ge 0
    \]

\end{proof}


\begin{theorem}[Hahn Decomposition]
\label{theorem: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:
    \begin{enumerate}
        \item $P$ is positive.
        \item $N$ is negative.
        \item $X = P \sqcup N$.
        \item[(U)] For any $P', N' \in \cm$ satisfying (1)-(3), $P \Delta P'$ and $N \Delta N'$ are null. 
    \end{enumerate}

    The disjoint union $X = P \sqcup N$ is the \textbf{Hahn decomposition} of $\mu$.
\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 3.3]{Folland}}}. ]
    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 \autoref{lemma:positive-sets-properties}, 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
    \[
    m(N) = \sup\bracs{\mu(A)|A \in \cm, A \subset N} > 0
    \]

    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
    \begin{enumerate}
        \item[(i)] For each $1 \le k \le n$, $M_k = m(A_k) - \mu(A_k) > 0$.
        \item[(ii)] For each $1 \le k \le n - 1$, $A_{k+1} \subset A_k$ with $\mu(A_{k+1}) - \mu(A_k) > M_{k}/2$.
    \end\{enumerate\}

    
    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
    \[
    M_{n+1} = m(A_{n+1}) - \mu(A_{n+1}) > 0
    \]

    Let $A = \bigcap_{n \in \natp}A_n$, then since $0 < \mu(A_2) < \infty$, $\mu(A) = \limv{n}\mu(A_n)$ by \autoref{proposition:signed-measure-properties}, so
    \[
    \frac{1}{2}\sum_{n = 1}^\infty M_n \le \sum_{n = 1}^{\infty}\mu(A_{n+1}) - \mu(A_n) = \mu(A) - \mu(A_1) < \infty
    \]
    and $M_n \to 0$ as $n \to \infty$. 
    
    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
    \[
    \mu(B) - \mu(A_n) > \mu(B) - \mu(A) > M_n = m(A_n) - \mu(A_n)
    \]

    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 $X = P \sqcup N = P' \sqcup N'$, 
    \[
    P \Delta P' = P \setminus P' \sqcup P' \setminus P = P \cap N' \sqcup P' \cap N
    \]

    is a union of two null sets, which is null. Likewise, 
    \[
    N \Delta N' = N \cap P' \sqcup N' \cap P
    \]

    is also a null set. 
\end{proof}

\begin{corollary}
\label{corollary:measure-bounded}
    Let $(X, \cm)$ be a measurable space, then
    \begin{enumerate}
        \item For any signed measure $\mu: \cm \to (-\infty, \infty)$, $\mu$ is bounded.
        \item For any complex measure $\mu: \cm \to \complex$, $\mu$ is bounded.
    \end{enumerate}
\end{corollary}
\begin{proof}
    (1): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem: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. 
\end{proof}

\begin{theorem}[Jordan Decomposition]
\label{theorem: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
    \begin{enumerate}
        \item $\mu^+ \perp \mu^-$.
        \item $\mu = \mu^+ - \mu^-$.
        \item[(U)] For any other pair $(\nu^+, \nu^-)$ satisfying (1) and (2), $\mu^+ = \nu^+$ and $\mu^- = \nu^-$.
    \end\{enumerate\}

    
\end{theorem}
\begin{proof}
    (1), (2): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$. For each $E \in \cm$, let
    \[
    \mu^+(E) = \mu(E \cap P) \quad \mu^-(E) = -\mu(E \cap N)
    \]

    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 $E \in \cm$,
    \[
    \mu^+(E) = \mu(E \cap P) = \mu(E \cap P') = \nu^+(E)
    \]

    Likewise,
    \[
    \mu^-(E) = -\mu(E \cap N) = -\mu(E \cap N') = \nu^-(E)
    \]
\end{proof}

