Let
- $X$ be a topological space,
- $\mathcal C_b(X)$ the space of real-valued bounded continuous functions on $X$,
- $\mathcal C_c(X)$ the space of real-valued continuous functions on $X$ with compact supports.
- $\mathcal M (X)$ the space of all complex Borel measures on $X$,
- $\mathcal M_{\mathbb R}(X)$ the space of all finite signed Borel measures on $X$, and
- $\mathcal M_+ (X)$ the space of all finite non-negative Borel measures on $X$.
For $\nu \in \mathcal M(X)$, let $|\nu| \in \mathcal M_+ (X)$ be its variation. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$ defined by $[\nu] := |\nu|(X)$. Then $(\mathcal M(X), [\cdot])$ is a Banach space.
- A subset $M$ of $\mathcal M(X)$ is said to be tight if for every $\varepsilon>0$ there is a compact subset $K$ of $X$ such that $\sup_{\nu \in M}| \nu| (X \setminus K) <\varepsilon$.
- A subset $M$ of $\mathcal M(X)$ is said to be bounded if $\sup_{\nu \in M} [\nu] <\infty$.
Let $\mu_n,\mu \in \mathcal M(X)$. We define
- weak convergence by $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X), $$
- weak$^*$ convergence by $$ \mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_c (X). $$
Previously, I proved that
Theorem Let $X$ be a locally compact separable metric space and $\mu, \mu_n \in \mathcal M_\mathbb R (X)$ such that $\limsup_n [\mu_n] \le [\mu]$. Then $\mu_n \rightharpoonup \mu$ if and only if $\mu_n \overset{*}{\rightharpoonup} \mu$.
Now I want to extend above theorem to $\mathcal M(X)$, i.e.,
Theorem: Let $X$ be a locally compact separable metric space and $\mu, \mu_n \in \mathcal M (X)$ such that $\limsup_n [\mu_n] \le [\mu]$. Then $\mu_n \rightharpoonup \mu$ if and only if $\mu_n \overset{*}{\rightharpoonup} \mu$.
The proof is essentially the same as the original one. However, I write it down and hope that someone helps me verify it. It's likely that I made some subtle mistakes in the transition from finite signed measures to complex measures.
Could you please have a check on my below attempt? Thank you so much!
Proof: One direction is obvious. Let's prove the reverse. We need the following lemmas, i.e.
Lemma 1 Let $X$ be a locally compact separable metric space. Then $X$ is a Radon space.
Lemma 2 Let $X$ be a locally compact Hausdorff space. Let $K$ be a compact subset of $X$ and $U$ an open subset of $X$ such that $K \subset U$. Then there is an open subset $V$ of $X$ such that $K \subset V \subset \overline V \subset U$ and $\overline V$ is compact.
Lemma 3 Let $X$ be a locally compact normal Hausdorff topological space. Let $\mu_n, \mu \in \mathcal{M}(X)$ such that $\mu_n \overset{*}{\rightharpoonup} \mu$. Then for any open subset set $O$ of $X$, $$ |\mu|(O) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(O) . $$
Lemma 4 Let $X$ be a topological space and $x, x_n \in X$. If every subsequence of $(x_n)$ has a further subsequence which converges to $x$. Then the sequence $(x_n)$ converges to $x$.
Prokhorov's theorem Let $X$ be completely regular topological space and $M$ a subset of $\mathcal M (X)$. If $M$ tight and bounded, then the closure of $M$ in $\sigma(\mathcal M (X), \mathcal C_b(X))$ is sequentially compact.
- $(\mu_n)$ is tight.
By Lemma 1, $\mu$ is tight. Fix $\varepsilon>0$. There is a compact subset $K$ of $X$ such that $|\mu| (K^c) <\varepsilon$. By Lemma 2, there is an open subset $O$ of $X$ such that $K \subset O$ and $K_\varepsilon := \overline O$ is compact. We have $$ \begin{align} \limsup_n |\mu_n|(X \setminus K_\varepsilon) &= \limsup_n \big [ [\mu_n]- |\mu_n|(K_\varepsilon) \big ] \\ &\le \limsup_n \big [ [\mu_n]- |\mu_n|(O) \big ] \\ &\le \limsup_n [\mu_n]- \liminf_n |\mu_n|(O) \\ &\le [\mu] - \liminf_n |\mu_n|(O). \end{align} $$
By Lemma 3, $$ \limsup_n |\mu_n|(X \setminus K_\varepsilon) \le [\mu] - |\mu|(O) = |\mu|(X \setminus O) \le |\mu| (X \setminus K) <\varepsilon. $$
- $\mu_n \rightharpoonup \mu$.
Let $\lambda$ be a subsequence of $\mathbb N$. By Prokhorov's theorem, there is a subsequence $\eta$ of $\lambda$ and $\hat \mu \in \mathcal M(X)$ such that $\mu_{\eta (n)} \rightharpoonup \hat \mu$ and thus $\mu_{\eta (n)} \overset{*}{\rightharpoonup} \hat \mu$ as $n \to \infty$. Notice that the weak$^*$ topology $\sigma (\mathcal M(X), \mathcal C_c(X))$ is Hausdorff, so $\hat \mu = \mu$. The claim then follows from Lemma 4.
