Question on exchanging probability with limit

Question

Consider a sequence of random variables $\{X_n\}_{n=1}^\infty$ where each random variable takes values in a set $\mathcal{X}$. Let $A \subset \mathcal{X}$ and assume that $X_n\rightarrow \alpha$ almost surely, where $\alpha \notin A$. From the convergence of $X_n$ we have that \begin{equation} \mathbb{P} (\lim_{n \rightarrow \infty} X_n \in A) =0. \end{equation} Under which conditions does this relation imply \begin{equation} \lim_{n \rightarrow \infty}\mathbb{P} ( X_n \in A) =0. \end{equation} I think the result should follow directly by an application of the Dominated Convergence theorem. I tried to use the theorem by writing the probability as the expected value of an indicator function, so we have: \begin{equation} \lim_{n \rightarrow \infty}\mathbb{P} ( X_n \in A) = \mathbb{E}[\lim_{n\rightarrow \infty} \mathbb{1}_{\{X_n \in A\}}]. \end{equation} However, I am not if I can bring the limit inside the indicator somehow to prove the result. Any help would be appreciated.

Answer 1

Since you are talking about $\lim_{n\to\infty}X_n$, I will assume $\mathcal X$ is a topological space.

A sufficient condition is to have $\alpha$ be a point in the interior of $A^c=\mathcal X\setminus A$. This means there is a neighborhood $U$ such that $\alpha \in U$ and $U\cap A=\varnothing$. Now, let $E_n$ be the event $\bigcap_{m\ge n} \{X_m\in U\}$, so $E_1\subset E_2\subset \dots$. Since $X_n\to \alpha$ a.s, we have $P(\bigcup E_n)=1$, which means $\lim_n P(E_n)=1$, which a fortiori implies $\lim_n P(X_n\notin A)=1$.

Edit: More detail. In a topological space, $x_n\to \alpha$ iff for every open set $U$ containing alpha, $x_n$ is "eventually" within alpha. This means there is an integer $n$ so $m\ge n$ implies $x_m\in U$. When $X_n$ is random, the property $\{X_m\in U\text{ for all } m\ge n\}$ is an event, which we denote $E_n$.

Since $X_n\to \alpha$ occurs almost surely, and convergence occurs implies at least one event $E_n$ occurs, we have $P(\bigcup E_n)=1$. Now, it is a standard result in probability theory that whenever $E_n$ is an increasing sequence of events, so that $E_1\subset E_2\subset \dots$, we have that $\lim_{n\to\infty} P(E_n)=P(\bigcup E_n)=1$.

Finally, consider the event $F_n = \{X_n\notin A\}$. Note that $F_n$ is implied by $E_n$, since $E_n\implies X_n\in U$ and $U$ is disjoint from $A$. Therefore, $\lim_{n\to\infty} P(E_n)=1$ implies $\lim_{n\to\infty} P(F_n)=1$ implies $\lim_{n\to\infty}P(X_n\in A)\to 0$.

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However, anything can happen if $\alpha$ is a limit point. For example, letting $\mathcal X=\mathbb R$, $A=(0,1)$, and $\alpha=0$, we have the following two possible situations: $$ X_n\sim \text{Unif}(-1/n,1/n),\quad\qquad\qquad P(X_n\in A)=1/2\not\to 0\\ Y_n\sim \text{Unif}(-1/n+1/{n^2},1/n^2),\qquad P(Y_n\in A)=1/n\to 0 $$

Finally, you cannot use Dominated convergence because you have not proved that that the random variables ${\mathbb 1}(X_n \in A)$ even converge almost surely. That is, $\lim_{n\to\infty}{\mathbb 1}(X_n \in A)$ does not exist almost surely.