How to prove that $\mu_n \rightharpoonup \mu$ IFF $\mu_n \overset{*}{\rightharpoonup} \mu$ and $\mu_n (X) \to \mu (X)$?

Question

Let

• $X$ be a metric space,
• $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
• $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions,
• $\mathcal C_0(X)$ be the space of real-valued continuous functions that vanish at infinity, and
• $\mathcal C_c(X)$ the space of real-valued continuous functions with compact supports.

Then $\mathcal C_b(X)$ and $\mathcal C_0(X)$ are real Banach space with supremum norm $\|\cdot\|_\infty$. Let $\mu_n,\mu \in \mathcal M(X)$. We define weak convergence by $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X), $$ and weak$^*$ convergence by $$ \mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_c (X). $$

Below "not-too-hard" theorem is mentioned in this thread, i.e.,

Theorem: $\mu_n \rightharpoonup \mu$ if and only if $\mu_n \overset{*}{\rightharpoonup} \mu$ and $\mu_n (X) \to \mu (X)$.

I'm trying to prove it, but I'm stuck at showing $$ \lim_m\lim_n \int_X (f-f_m) \mathrm d \mu_n = 0. $$

Could you elaborate on how to finish the proof?

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My attempt: One direction is obvious. Let's prove the reverse. Assume $\mu_n \overset{*}{\rightharpoonup} \mu$ and $\mu_n (X) \to \mu (X)$. Fix $f \in \mathcal C_b (X)$ and $\varepsilon>0$. Let $(\mu^+, \mu^-)$ with $\mu = \mu^+ - \mu^-$ be the Jordan decomposition of $\mu$. Let $|\mu| := \mu^+ + \mu^-$. By definition, $$ \int_X f \mathrm d \mu := \int_X f \mathrm d \mu^+ - \int_X f \mathrm d \mu^-. $$

Notice that $\mathcal C_c(X)$ is dense in $(L_1 (|\mu|), \|\cdot\|_{L_1(|\mu|)})$, so there is a sequence $(f_m) \subset L_1 (\mu|)$ such that $\|f_m - f\|_{L_1(|\mu|)} \to 0$, i.e., $$ \int_X |f_m-f| \mathrm d |\mu| \to 0 \quad \text{as} \quad m \to \infty. $$

Notice that $$ \begin{align} \left | \int_X (f_m-f) \mathrm d \mu \right | &= \left | \int_X (f_m-f) \mathrm d \mu^+ - \int_X (f_m-f) \mathrm d \mu^- \right | \\ &\le \int_X |f_m-f| \mathrm d \mu^+ + \int_X |f_m-f| \mathrm d \mu^- \\ &= \int_X |f_m-f| \mathrm d |\mu| . \end{align} $$

This implies $$ \int_X f_m \mathrm d \mu \to \int_X f \mathrm d \mu \quad \text{as} \quad m \to \infty. $$

We have a decomposition $$ \int_X f \mathrm d (\mu_n-\mu) = \int_X (f_m-f) \mathrm d \mu + \int_X f_m \mathrm d (\mu_n-\mu) + \int_X (f-f_m) \mathrm d \mu_n. $$

Answer 1

This is in response to a comment by the OP. It is the version of the Theorem in his posting under the additional assumption that $X$ is a metric space:

Theorem A: Suppose $(S,d)$ is a locally compact separable metric space. Let $(\mu_n,\mu)$ a sequence of finite nonnegative measures. The following statement are equivalent:

1. $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu$ (convergence in the topology $\sigma(\mathcal{M}(S),\mathcal{C}_b(S))$)
2. $\mu_n\stackrel{v}{\longrightarrow}\mu$ as $n\rightarrow\infty$ (convergence in the topology $\sigma(\mathcal{M}(S),\mathcal{C}_{00}(S))$, and $\mu_n(S)\xrightarrow{n\rightarrow\infty}\mu(S)$.

That (1) implies (2) is obvious.

The proof that (2) implies (1) is based in a well known result:

Theorem L: Let $(S,d)$ be a metric space. For any net $\{\mu_\alpha:\alpha\in D\}\subset\mathcal{M}^+(S)$ and $\mu\in \mathcal{M}^+(S)$,

• $\mu_\alpha\Rightarrow\mu$ if and only if \begin{align} \liminf_\alpha\int f\,d\mu_\alpha\geq \int f\,d\mu\tag{1}\label{one} \end{align} for all $f\in L_b(S)$.
  If in addition $(S,d)$ is a locally compact separable metric space,
• If $\mu_\alpha\stackrel{v}{\longrightarrow} \mu$, then \eqref{one} holds for all $0\leq f\in L_b(S)$.

Here, $L_b(S)$ the set of all lower semicontinuous functions that are bounded below.

Let $f\in L_b(S)$ with $c\leq f$ for some constant $c$. Then $0\leq f-c\in L_b(S)$ and by the second part of Theorem L, $\liminf_n\int (f-c)\,d\mu_n\geq \int (f-c)\,d\mu$. The assumption $\mu_n(S)\rightarrow\mu(S)$ implies that \begin{align} \liminf_n\int f\,d\mu_n\geq \int f\,d\mu . \end{align} The implication $(2)\Rightarrow(1)$ in Theorem A follows from the first part of Theorem L.

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Here is a proof of Theorem L.

Suppose that $\mu_\alpha\Rightarrow\mu$ and let $g\in L_b(S)$ with $g\geq c$. There is a sequence $g_k$ of bounded Lipschitz functions such that $c\leq g_k\leq g_{k+1}\nearrow g$. Hence, for each $k$ \begin{align} \liminf_\alpha\int g\,d\mu_\alpha\geq \liminf_\alpha \int g_k\,d\mu_\alpha =\int g_k\,d\mu. \end{align} As $\mu(S)<\infty$, $\liminf_\alpha\int g\,d\mu_\alpha\geq \int g\,d\mu$ by monotone convergence.

Conversely, suppose $f\in\mathcal{C}_b(S)$. Since $\mathcal{C}_b(S)\subset L_b(S)$, both $f$ and $-f$ are in $L_b(S)$, so \begin{align} \liminf_\alpha\int f\,d\mu_\alpha&\geq \int f\,d\mu\\ \liminf_\alpha\int -f\,d\mu_\alpha&\geq \int -f\,d\mu \end{align} Therefore, $\lim_\alpha \int f\,d\mu_\alpha =\int f\,d\mu$.

For the last statement, let $0\leq f\in L_b(S)$ and let $f_k\in C_b(S)$ be such that $0\leq f_k\nearrow f$ pointwise. Since $S$ is locally compact and separable, there is a sequence of open sets $V_j$ with compact closure such that $\overline{V}_j\subset V_{j+1}\nearrow S$. Choose $v_j\in C_{00}(S)$ so that $\mathbb{1}_{\overline{V}_j}\leq v_j\leq \mathbb{1}_{V_{j+1}}$ and $\operatorname{supp}(v_j)\subset V_{j+1}$. Let $f_{kj}=f_kv_j$; clearly $f_{kj}\in C_{00}(S)$ and $f_{kj}\nearrow f_k$ as $j\nearrow\infty$. Then for all $k$ and $j$ \begin{align} \liminf_\alpha\int f\,d\mu_\alpha\geq \liminf_\alpha\int f_k\,d\mu_\alpha\geq \liminf_\alpha\int f_{kj}\,d\mu_\alpha=\int f_{kj}\,d\mu \end{align} The result follows now by monotone convergence by letting $j\nearrow\infty$ and then $k\nearrow\infty$.

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All the arguments presented here depend heavily on the metrizability of $S$. I don't think they can be generalize easily into the more general setting where $(S,\tau)$ is a l.c.H space.

Answer 2

For $\nu \in \mathcal M(X)$, let $(\nu^+, \nu^-)$ be its Jordan decomposition, and $|\nu| := \nu^+ + \nu^-$ 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 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 bounded if $\sup_{\nu \in M} [\nu] <\infty$.

We are going to prove below generalization, 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 $\mu_n \overset{*}{\rightharpoonup} \mu$.

All the ideas come from this paper.

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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.

1. $\lim_n [\mu_n] = [\mu]$.

Consider the map $$ L:\mathcal M(X) \to \mathcal C_c(X)^*, \nu \mapsto \left (L_\nu :f \mapsto \int_X f \mathrm d \nu \right). $$

Then $L$ 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 $\sigma(\mathcal C_c(X)^*, \mathcal C_c(X))$. So $\|L_\mu| \le \liminf_n \|L_{\mu_n}\|$ and thus $[\mu] \le \liminf_n [\mu_n]$.

2. $(\mu_n)$ 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 $$ \begin{align} \limsup_n |\mu_n|(X \setminus K_\varepsilon) &= \limsup_n \big [ [\mu_n]- |\mu_n|(K_\varepsilon) \big ] \\ &\le \limsup_n \big [ [\mu_n]- |\mu_n|(O) \big ] \\ &= [\mu] - \liminf_n |\mu_n|(O). \end{align} $$

By Lemma 3, $$ \limsup_n |\mu_n|(X \setminus K_\varepsilon) \le [\mu] - |\mu|(O) = |\mu|(X \setminus O) \le |\mu| (X \setminus K) <\varepsilon. $$

3. $\mu_n \rightharpoonup \mu$.

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.