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Let $M$ be a compact metric space and $\mathcal{B}$ its Borelian $\sigma$-algebra. Consider $\{\mu_{n}\}_{n\in\mathbb N}$ as a sequence of Borelian probabilities on $M$. Suppose that there exists a Borelian probability $\mu$ on $M$ and a generating algebra $\mathcal{A}$ (i.e. $\mathcal A$ is an algebra and $\sigma(\mathcal{A}) = \mathcal B$) such that $$\mu_n(A)\longrightarrow \mu(A),\ \forall\ A\in \mathcal A\ \text{and}\ \mu(\partial A)=0,\ \forall \ A\in\mathcal{A}. \quad \quad (*)$$

I would like to know if $(*)$ implies that $\mu_n\to\mu$ in the weak* topology, i.e. for every continuous function $f: M \to \mathbb R$ $$\int_M f\ \text{d}\mu_n \longrightarrow \int_M f\ \text{d}\mu. $$

My attempt

I tried to use the monotone class theorem for functions. I defined the set $$\mathcal H:=\left\{f:M\to\mathbb R;\ f \text{ is bounded, measurable and }\int_M f\ \text{d}\mu_n \longrightarrow \int_M f\ \text{d}\mu\right\}. $$

So if we prove that

  1. if $A\in \mathcal A\Rightarrow$ $1_A \in \mathcal H,$
  2. if $f,g\in\mathcal H$ $\Rightarrow$ $f+cg \in\mathcal{H}$, for any real number $c$,
  3. if $f_n \in \mathcal{H}$ is a sequence of non-negative functions that increase to a bounded function $f$ $\Rightarrow$ $f \in \mathcal{H}$,

holds then, by the monotone class theorem for functions, $\mathcal H$ will all the bounded measurable functions, and we are done. The conditions $1$ and $2$ are obvious to be checked. However, I was not able to conclude the last condition.

Can anyone help me?

Edit: I was thinking and this approach does not make sense since the condition that I am trying to check is a way stronger than convergence in the weak* topology.

mathworker21
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Matheus Manzatto
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    You must search for Portmanteau Theorem – Eduardo Mar 27 '20 at 00:11
  • One of the implications of Portmanteau Theorem is that $\mu_{n}\to \mu$ $ \Leftrightarrow$ $\mu_n(B) \to \mu(B)$ $\forall$ $B\in \mathcal{B}$ such that $\mu(\partial B) = 0$. The problem here is the fact that $\mu_n(A) \to \mu(A)$ $\forall$ $A$ in a generating Algebra. I don't know how to overcome this issue. – Matheus Manzatto Mar 27 '20 at 00:19
  • By a Borelian probability measure $\mu$, do you mean just a probability measure on the measurable space $(M,\mathcal{B}(M))$ ? Is there any regularity imposed on $\mu$ ? For example, for each $A\in\mathcal{B}(M)$, $\mu(A) = \inf {\mu(U)\mid A\subseteq U\mbox{ and }U\mbox{ is an open subset of M.}}$? – Danny Pak-Keung Chan Mar 27 '20 at 00:19
  • Also, inner regularity? – Danny Pak-Keung Chan Mar 27 '20 at 00:20
  • Since $M$ is a metric space, all probabilities are regular, i.e. for every measurable set $B$ and $\varepsilon>0$ there exist a closed set $F$ and an open set $A$ such that $F\subset B\subset A$ and $\mu(A\setminus F)<\varepsilon$. Since $M$ is compact, then $F$ is compact as well. – Matheus Manzatto Mar 27 '20 at 00:26
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    If $\mathcal{A}$ is a generating algebra you can approximate the measure of every borelian by an element of $\mathcal{A}$. I think that this plus an $\epsilon/3$ argument do the job – Eduardo Mar 27 '20 at 00:32
  • @Eduardo I have never approximated a borelian set using a generating algebra. Can you please explain to me how I do this? (Or suggest some result) – Matheus Manzatto Mar 27 '20 at 00:39
  • If ${\cal A}$ is a generating algebra then for every $B\in \sigma({\cal A})$ and $\epsilon>0$ there exists an element $A\in {\cal A}$ such that $\mu(A\Delta B)<\epsilon$. This is in part consequence of Caratheodoy Extension Theorem – Eduardo Mar 27 '20 at 00:46
  • @Eduardo I do not think that this will work... since we can not find an element $A\in\mathcal{A}$ such that $\mu_n(A \Delta B) <\varepsilon$ $\forall$ $n$. I think this $A$ can not be taken uniformly. – Matheus Manzatto Mar 27 '20 at 01:11
  • @MatheusManzatto why can't you just do the proof of Portmanteau's theorem, specifically the implication $\mu_n(B) \to \mu(B)$ for all continuous $B$ --> $\int fd\mu_n \to \int fd\mu$ for all continuous $f$, but just use the algebra $\mathcal{A}$ you have? The proof of that Portmanteau implication is to first prove $\int fd\mu_n \to \int fd\mu$ holds for all simple functions and then do some standard approximations, but you can approximate any continuous function with simple functions that are $\mathcal{A}$-measurable, since $\sigma(\mathcal{A}) = \mathcal{B}$. Am I missing something? – mathworker21 Mar 29 '20 at 05:34
  • @mathworker21 This is not the proof that I know (the one that I know is from the book "Foundations of Ergodic Theory - Marcelo Viana" (Theorem 2.1.2) ). Moreover, I am not sure that you can approximate any continuous function with simple functions of $\mathcal A$ (do you have any results of this to recommend me?). – Matheus Manzatto Mar 29 '20 at 17:52

1 Answers1

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Here is a proof if $\mathcal{A}$ consists only of open sets or consists only of closed sets.

First suppose $\mathcal{A}$ consists only of open sets. Let $\nu$ be a weak* limit of $(\mu_n)_n$. Then $\nu(A) \le \liminf_n \mu_n(A) = \mu(A)$ for all $A \in \mathcal{A}$. So $\nu(B) \le \mu(B)$ for all $B \in \mathcal{B}$. Therefore, $\nu = \mu$ (since both are probability measures). Closed sets case is similar.

The statement is false if $M$ is not compact. Take, for example, $M = \mathbb{R}$ and $\mathcal{A} = \{[x,y] : -\infty < x < y < +\infty\}$, the set of compact intervals. Let $\mu_n = 1_{[n,n+1]}$. Then $\mu_n(A) \to 0$ for each $A \in \mathcal{A}$, but $\mu_n$ does not converge weak* to $0$, since, for example, $\int_{\mathbb{R}} 1d\mu_n = 1$ for each $n$.

Consequently, the approach you outlined in your question probably won't work. It seems that one has to work with a weak* limit of $(\mu_n)_n$, as (partially) done above.

mathworker21
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  • How do you know that every $A\in\mathcal A$ is a continuity set of $\nu$? Because, $\nu(A) = \mu(A)$ only if $\nu(\partial A) = 0$, right? – Matheus Manzatto Mar 30 '20 at 01:55
  • @MatheusManzatto gahhhhhhhh. you're right. well, i don't know if you can say "$\nu(A) = \mu(A)$ only if $\nu(\partial A) = 0$", but we can only immediately deduce $\nu(A) = \mu(A)$ if $\nu(\partial A) = 0$. thanks for the catch – mathworker21 Mar 30 '20 at 02:01
  • @MatheusManzatto Do you know if the result is true if $M$ is non-compact? I realized that working with a weak* limit of $(\mu_n)_n$ seems like the only way to really use compactness; all the other ideas I had in mind wouldn't use compactness. – mathworker21 Mar 30 '20 at 02:01
  • @MatheusManzatto wait, no. you tricked me just now. you said in your problem that $\mu_n(A) \to \mu(A)$ for all $A$. you also said that $\mu(\partial A) = 0$ for all $A$. So the answer to the first question in your comment is "because you said so". Maybe you have to change/clarify your problem statement? – mathworker21 Mar 30 '20 at 02:02
  • First of all, I don't know if the result is correct in the case ($M$ compact), let alone if it is valid if $M$ is non-compact. In my problem, I know that there exists a probability $\mu$ such that $\mu_n(A) \rightarrow \mu(A)$ for all $A\in\mathcal A$. What I am point is that we do not know, a priori, that $\mu_n(A)\to \nu(A)$, because we do not know that $\nu(\partial A) = 0$ (we only know that $\mu(\partial A) = 0$). – Matheus Manzatto Mar 30 '20 at 02:08
  • @MatheusManzatto I was more directly asking if you found a counterexample when $M$ is non-compact. I think knowing if a counterexample exists would be very important, since then if the result is true for $M$ compact, we'd need to look at a weak* limit of $(\mu_n)_n$ probably. Also, sorry for being so slow; I now get what you're saying now re the boundary issue. – mathworker21 Mar 30 '20 at 02:11
  • We do not need to be sorry it is okay :). I also was not able to find a counterexample in the non-compact case. But your idea was really good, I will think about whether it is possible to overcome this problem. – Matheus Manzatto Mar 30 '20 at 02:13
  • @MatheusManzatto Well, I think the continuity sets should generate $\mathcal{A}$. Does this finish things off? https://math.stackexchange.com/questions/1634436/sigma-field-generated-by-the-continuity-sets-of-a-measure – mathworker21 Mar 30 '20 at 02:20
  • I think it does not finish things off since we do not choose $\mathcal A$. But I really liked your answer. If there is no better answer (the general case) until the end of the bounty I will give you 200 points. – Matheus Manzatto Mar 30 '20 at 03:14
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    @MatheusManzatto See the updated answer. Also, I wasn't trying to say that we can choose $\mathcal{A}$. I think the following general statement should be true, and it is clearly sufficient to finish off the problem: If $M$ is a compact metric space, $\mathcal{A}$ generates the Borel sigma-algebra, and $\nu$ is a probability measure, then ${A \in \mathcal{A} : \nu(\partial A) = 0}$ also generates the Borel sigma-algebra. – mathworker21 Mar 30 '20 at 03:33
  • Now I see how you want to finish the question (sorry, my fault). I will try to prove that ${A∈ \mathcal A:ν(∂A)=0}$ also generates the Borel sigma-algebra which makes sense to be true. I have seen your updated answer, you counter-example was superb. – Matheus Manzatto Mar 30 '20 at 03:57
  • @MatheusManzatto Hi. Any progress on showing $\sigma\left({A \in \mathcal{A} : \nu(\partial A) = 0}\right) = \mathcal{B}$? Should I think about it? – mathworker21 Apr 01 '20 at 07:33
  • I was not able to prove such equality, but I would be very happy if that were true. Since it would prove my question in all generality. But as I said, your progress ( in case $\mathcal{A}$ consists of only closed or open sets) has already been considerable progress, and as I work in general considering the sigma-algebra generated by intervals (or rectangles), this ends the question for practical purposes, but not in all its generality. – Matheus Manzatto Apr 01 '20 at 20:47