A (sort of) converse to the monotone class theorem

Question

In my notes we have the monotone class theorem which says that if a vector space of functions contains all the indicator functions of measurable sets and is closed under bounded monotone limits then it contains all bounded monotone functions.

We also have a definition of weak convergence of probability measures as $\mu_n \to \mu$ if for every bounded continuous $g$ we have $\mu_n(g) \to \mu(g)$. My question is does this imply that we have $\mu_n(1_A) \to \mu(1_A)$ for all measurable sets $A$ where $1_A$ is the indicator function of $A$? It seems like it should be trivial but $1_A$ is not a continuous function so we can't just use the definition.

I think it should be true but I can't come up with a proof. If the domain of the functions was $\mathbb{R}$ I belive it should be possible to use prove this result for all $A$ of the form $(-\infty,x]$ using dominated convergence and then using Dynkin's lemma to extend to indicators of all (Borel) measurable sets. However if the domain is not $\mathbb{R}$ I don't see how to do this. Alternatively if it is false what additional conditions do I need to make this true?

Thanks for your help!

Answer 1

No, to demand $\mu_n(A)\to\mu(A)$ for all measurable sets $A$ is too much.

You have seen the example $\delta_{1/n}\to\delta_0$ weakly in Gabriel's answer.

Here is another example, which is closer to probability theory: If $\mu_n$ is the "standardized" binomial distribution with parameters $n,\frac{1}{2}$ (i.e. we translate and scale so to have mean $0$ and variance $1$), then it converges weakly to the standard normal distribution. However, for every $n$ the support of $\mu_n$ is finite, so there is a countable set $A=\bigcup\operatorname{supp}\mu_n$ so that $\mu_n(A)=1$ for all $n$ and of course $\mu(A)=0$. (Of course there are other distribution family we could use instead of binomial(n,1/2), this is just the simplest example of central limit theorem.)

However, you have $\limsup\mu_n(F)\leq\mu(F)$ for all closed $F\subseteq S$ and dually $\liminf\mu_n(G)\geq\mu(G)$ for all open $G\subseteq S$, where $S$ is the underlying metric space. Hence if $A\in\mathcal{B}(S)$ is $\mu$-boundaryless (i.e., $\partial A=\overline{A}-A^\circ$ is $\mu$-null), then $\lim\mu_n(A)=\mu(A)$. In the example $\delta_{1/n}\to\delta_0$ the boundary of $A$ contains $\{0\}$, and in my example above, $\partial A=\overline{A}$ is actually the whole of $\mathbb{R}$.

Answer 2

This is false in general. Here's a counterexample in the context of probability measures: consider $X_n=\frac 1n $ and $A=[-1,0]$. $(X_n)$ converges almost surely to $0$, hence $P_{X_n}$ converges weakly to $\delta_{\{0\}}$. But $P_{X_n}(A)=0$, whereas $\delta_{\{0\}}(A)=1$.