Let
- $X$ be a metric space,
- $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
- $\mathcal M_+(X)$ the space of all finite nonnegative Borel measures on $X$, and
- $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$.
Then $\mathcal C_b(X)$ is a real Banach space with supremum norm $|\cdot|_\infty$. For $\mu \in \mathcal M(X)$, let $(\mu^+, \mu^-)$ be its Jordan decomposition and $|\mu| := \mu^+ + \mu^-$ its variation. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$ where $[\mu] := |\mu|(X)$. Then $(\mathcal M(X), [\cdot])$ is a Banach space. For $\mu_n,\mu \in \mathcal M(X)$, we define the weak convergence $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X). $$
I would like to prove below result, i.e.,
Theorem: Let $\mu_n,\mu \in \mathcal M(X)$ such that $\limsup_n [\mu_n] \le [\mu]$. Then $\mu_n \rightharpoonup \mu$ IFF $\mu^+_n \rightharpoonup \mu^+$ and $\mu^-_n \rightharpoonup \mu^-$.
Could you have a check on my attempt?
Proof: The reverse direction is obvious. Let's prove the other one.
First, we prove that $\liminf_n [\mu_n] \ge [\mu]$. It's well-known that $$ \mathcal M(X) \to \mathcal C_b(X)^*, \nu \mapsto \left (L_\nu :f \mapsto \int_X f \mathrm d \nu \right). $$ is an isometrically isomorphic embedding. This implies $[\nu] = \|L_\nu\|$ for all $\nu \in \mathcal M(X)$. Notice that $\mu_n \rightharpoonup \mu$ if and only if $L_{\mu_n} \to L_\mu$ in the weak$^*$ topology $\sigma(\mathcal C_b(X)^*, \mathcal C_b(X))$. As such, $\|L_\mu| \le \liminf_n \|L_{\mu_n}\|$ and thus $\liminf_n [\mu_n] \ge [\mu]$. As such, $\lim_n [\mu_n] = [\mu]$.
Lemma Let $\mu_n,\mu \in \mathcal M(X)$ such that $\mu_n \rightharpoonup \mu$ and $[\mu_n] \to [\mu]$, then $|\mu_n| \rightharpoonup |\mu|$.
By our Lemma, $|\mu_n| \rightharpoonup |\mu|$. So we have $\mu^+_n - \mu_n^- \rightharpoonup \mu^+ - \mu^-$ and $\mu^+_n + \mu_n^- \rightharpoonup \mu^+ + \mu^-$. The claim then follows.