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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.

Analyst
  • 6,351
  • The dual of $\mathcal{C}_b(X)$ is much larger than $\mathcal{M}(X)$, it contains all normal chargers. You map can't be an isomorphisms. The result you are interested seems to be a consequence of the Prohorov theorem provided some tightness conditions hold. – Mittens Nov 06 '22 at 14:24
  • @OliverDíaz I only use that the map is an embedding... – Analyst Nov 06 '22 at 14:26
  • Still, tightens seems to be needed in your Lemma, or at least, implied from the conditions on your Lemma. That is what allows things to work as in the Riesz representation theorem. – Mittens Nov 06 '22 at 14:29
  • @OliverDíaz my version works in a general metric space, so i think Riesz theorem is not applicable here. – Analyst Nov 06 '22 at 16:41
  • I am not saying that the Riesz representation has to be applied. Tighness is needed in many of this arguments. You have not shown any tightness. – Mittens Nov 06 '22 at 16:52
  • @OliverDíaz the essential of the proof is this lemma. I really hope that you can have a check on it. – Analyst Nov 06 '22 at 17:03

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