I'm trying to grasp the idea of Hahn decomposition by proving below result. Could you verify if my attempt is fine?
Let $X$ be a topological space and $\mathcal M:=\mathcal M(X)$ the space of all finite signed Borel measures on $X$. By Jordan decomposition, each $\mu \in \mathcal M$ can be written uniquely as $\mu = \mu^+ – \mu^-$ such that
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$\mu^+, \mu^- \in \mathcal M$ are non-negative, and
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for any Hahn decomposition (h.d.) $(P, N)$ of $\mu$ one has for all $B \in \mathcal B(X)$ that $\mu^+(B)=0$ if $B \subset N$ and that $\mu^-(B)=0$ if $B \subset P$. Here the first (resp. second) coordinate of the pair $(P, N)$ is the positive (resp. negative) region of $\mu$.
We define a norm $\| \cdot \|$ on $\mathcal M$ by
$$
\|\mu\| := \mu^+ (X) + \mu^- (X).
$$
Theorem: $(\mathcal M, \|\cdot\|)$ is a Banach space.
I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.
Best Answer
Let's first prove that $\|\cdot\|$ is indeed a norm.
Notice that if $(P, N)$ is a h.d. of $\mu$, then $(P, N)$ is a h.d. of $c \mu$ for all $c \ge 0$. Also, if $(P, N)$ is a Hahn decomposition of $\mu$, then $(N, P)$ is a h.d. of $c \mu$ for all $c \le 0$. Let $\mu = \mu^+ - \mu^-$. It's easy to see that $(c\mu)^+ = c\mu^+$ and $(c\mu)^- = c\mu^-$ for all $c \ge 0$. Also, $(c\mu)^+ = |c|\mu^-$ and $(c\mu)^- = |c|\mu^+$ for all $c \le 0$. It follows that $\|c \mu\| = |c| \|\mu\|$ for all $c$.
If $\mu \equiv 0$, then $\mu^+ = \mu^- \equiv 0$. If $\|\mu\|=0$, then $\mu^+(X)=\mu^-(X)=0$ and thus $\mu^+ = \mu^- \equiv 0$. Hence $\mu \equiv 0 \iff \|\mu\|=0 \iff \mu^+ = \mu^- \equiv 0$.
Let $(P_\mu, N_\mu), (P_\nu, N_\nu), (P_{\mu+\nu}, N_{\mu+\nu})$ be h.d.'s of $\mu, \nu, \mu+\nu$ respectively. For any $B \in \mathcal B(X)$, we have $\mu (B \cap P_{\mu+\nu}) \le \mu (B \cap P_{\mu+\nu} \cap P_\mu) \le \mu(B\cap P_\mu)$ and $\mu (B \cap N_{\mu+\nu}) \ge \mu (B \cap N_{\mu+\nu} \cap N_{\mu}) \ge \mu(B\cap N_\mu)$. Similarly, $\nu (B \cap P_{\mu+\nu}) \le \nu(B\cap P_\nu)$ and $\nu (B \cap N_{\mu+\nu}) \ge \nu(B\cap N_\nu)$. Thus \begin{align} (\mu+\nu)^+(B) &= (\mu+\nu)(B \cap P_{\mu+\nu}) = \mu(B \cap P_{\mu+\nu}) + \nu(B \cap P_{\mu+\nu}) \\ &\le \mu(B\cap P_\mu)+\nu(B\cap P_\nu) = \mu^+(B) +\nu^+(B) \\ -(\mu+\nu)^-(B) &=(\mu+\nu)(B \cap N_{\mu+\nu}) = \mu(B \cap N_{\mu+\nu}) + \nu(B \cap N_{\mu+\nu}) \\ &\ge \mu(B\cap N_\mu)+\nu(B\cap N_\nu) = -\mu^-(B) -\nu^-(B). \end{align} It follows that $\|\mu+\nu\| \le \|\mu\|+\|\nu\|$.
Next we prove that $(\mathcal M, \| \cdot \|)$ is complete. Let $(\mu_n)$ be a Cauchy sequence in $\mathcal M$. We have $|\mu_n(B)-\mu_m(B)| = |(\mu_n-\mu_m)(B)| \le \|\mu_n-\mu_m\|$ for all $B \in \mathcal B(X)$. In particular, $(\mu_n (B))_n$ is a Cauchy sequence for all $B \in \mathcal B(X)$. We define $\mu \in \mathcal M$ by $\mu(B) := \lim_n \mu_n(B)$.
Fix $\varepsilon > 0$. There is $N$ such that $\|\mu_n-\mu_m\| < \varepsilon$ for all $m,n \ge N$. $$ |(\mu_n -\mu)(B)| =\lim_m |(\mu_n -\mu_m)(B)| \le \varepsilon \quad \forall B \in \mathcal B(X), \forall n\ge N. $$
Notice that $$ \mu^+(B) = \sup \{ \mu(C) \mid C\in \mathcal B(X), C \subset B\} \quad \text{and} \quad \mu^-(B) =\sup \{-\mu(C) \mid C\in \mathcal B(X), C \subset B\}. $$
It follows that $$ (\mu_n-\mu)^+ (X) = \sup \{ (\mu_n-\mu)(B) \mid B\in \mathcal B(X)\} \le \varepsilon \quad \forall n\ge N. $$
Similarly, $$ (\mu_n-\mu)^- (X) = \sup \{ -(\mu_n-\mu)(B) \mid B\in \mathcal B(X)\} \le \varepsilon \quad \forall n\ge N. $$
As a result, $\mu_n \to \mu$ in $\|\cdot\|$. This completes the proof.
We define another norm $\|\cdot\|_\infty$ on $\mathcal M$ by $$ \|\mu\|_\infty := \sup \{|\mu(B)| \mid B \in \mathcal B(X)\}. $$
Notice that $-\mu^-(B) \le \mu(B) \le \mu^+(B)$, so $|\mu(B)| \le \|\mu\|$. Then $$ \|\mu\|_\infty \le \|\mu\|. $$
Let $(P, N)$ be a h.d. of $\mu$. Then $\|\mu\|_\infty \ge \max \{|\mu(P)|, \mu(N)|\}$. Then $$ 2\|\mu\|_\infty \ge |\mu(P)|+|\mu(N)| = \mu^+(X)+\mu^-(X) = \|\mu\|. $$
It follows that $\|\cdot\|_\infty$ is equivalent to $\|\cdot\|$.