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

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?

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 $$\int_X f \mathrm d (\mu_n-\mu) = \int_X (f_m-f) \mathrm d \mu + \int_X f_m \mathrm d (\mu_n-\mu) + \int_X (f-f_m) \mathrm d \mu_n.$$

• I now the result holds of $X$ is a locally compact separable matrix space (for example Doob's measure theory, Chapter VIII, paragraph 1 or Betrsekas, D.P and Shere S.E. Stochastic Optima Control: The DiscreteTime Case, Academy Press, 1978). In general, I am not aware of that Commented Nov 1, 2022 at 20:55
• @OliverDíaz It seems the theorem you referred to is at page 140 of Doob's Measure Theory. However, there is a tightness condition. It seems to me the "not-too-hard" theorem mentioned here is not true. Could you please have a check on it? Commented Nov 3, 2022 at 19:12
• @OliverDíaz You meant that if $X$ is locally compact seprable and $\mu_n \overset{*}{\rightharpoonup} \mu$, then $(\mu_n)$ is uniformly tight, right? It seems from this recent question that [$\mu_n \overset{*}{\rightharpoonup} \mu$ and $\mu_n (X) \to \mu (X)$] is equivalent to [$\int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu$ for all continuous $f$ that is either constant or has a finite limit at infinity]. Btw, another approach is always welcome. Commented Nov 3, 2022 at 19:48
• @OliverDíaz Could you write up an answer that contains sketch of the proof? Then I can fill in the details later... Commented Nov 3, 2022 at 20:49
• I wrote a rather detail statement and proof of a version of the Theorem statement on your posting with the assumption that $S$ is a locally compact metric space. Commented Nov 3, 2022 at 21:54

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.

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 $$k\nearrow\infty$$.

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.

• First, I would like to thank you very much for your detailed answer. It's quite late on my side, so I will read it carefully tomorrow. I would like to confirm that with $X$ being locally compact separable metric space, we can drop the requirement $\mu_n (X) \to \mu (X)$, right? Commented Nov 3, 2022 at 22:18
• No, that is needed! otherwise you may have escape of mass to infinity. Commented Nov 3, 2022 at 22:25
• I asked because I do not see that assumption in your theorem... Commented Nov 3, 2022 at 22:26
• It seems we can generalize above theorem to finite signed Borel measures. Please see my answer below. Commented Nov 6, 2022 at 18:53

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.

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 $$[\mu] \le \liminf_n [\mu_n]$$.

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

1. $$\mu_n \rightharpoonup \mu$$.

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.

• (+1) as I mentioned earlier, the key was to get some tightness, which you managed to do. The rest is Prohorov's theorem. Commented Nov 6, 2022 at 19:15
• @OliverDíaz The key ingredient, which is Lemma 3, can be generalized for complex Borel measures. Please see here if you are interested. Commented Nov 7, 2022 at 17:44
• I will take a look later today, however I think it should be straight forward: $\frac{|a|+|b|}{\sqrt{2}}\leq\sqrt{a^2+b^2}\leq|a|+|b|$. Commented Nov 7, 2022 at 18:02
• For Polish spaces tightness and boundedness yields sequential compactness even for families of complex measures. just apply result to real and imaginary parts. Commented Nov 8, 2022 at 12:58
• I'll take a look at it, though I am busy at work and it might take me some time to digest the argument if they are somehow complicated. Commented Nov 9, 2022 at 2:35