Suppose $X_n \rightarrow X$ a.s. and for each $n$, $X_n \perp \textit F$. Then is it true that $X \perp \textit F$?

Question

For random variables $X_n$ and $X$, suppose $X_n \rightarrow X$ a.s. and for each $n$, $X_n \perp \mathcal F$ i.e. independent with $\mathcal F$.

My question is:

Is it true that $X \perp \mathcal F$ ?

I know that I can assume $X_n \rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $\sigma (X_n : n \geq1)$ measurable. By assumption, $\sigma (X_n) \perp \mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $\sigma (X_n : n \geq1)$ and is independent with $\mathcal F$ , but it gets me nowhere.

Any hint is appreciated.

Answer 1

We can follow these steps:

1. Using the dominated convergence theorem, show that for all continuous and bounded function $\varphi\colon\mathbb R\to\mathbb R$ and each $F\in\mathcal F$, the following equality holds: $$\tag{*} \mathbb E\left[\varphi(X)\mathbf 1_F\right]= \mathbb E\left[\varphi(X) \right]\mathbb P(F).$$

2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(\cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $\varphi$ is the indicator function of $S$.

3. This shows that $\mathbb P(\{X\in S\}\cap F)=\mathbb P(X\in S)\mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.