Does $\mathbb{P}$-a.s. convergence preserve independence?

Question

Let $\mathcal F$ be a $\sigma$-algebra and $X_n$ RV s.t. $X_n$ is independent of $\mathcal F$ for all $n$. Also let $X_n \to X$ $\mathbb{P}-$a.s.. Is $X$ independent of $\mathcal F$ now too?

Answer 1

Choose a random variable $Y$ such that $X=Y$ $\mathbf P$-a.s. and $X_n \to Y$ pointwise. As all $X_n$ are $\sigma(\{X_n : n \in \mathbf N\})$-measurable, $Y$ is also. By assumption $F$ and $\sigma(X_n: n \in \mathbf N)$ are independent, hence so are $F$ and $\sigma(Y) \subseteq \sigma(X_n: n \in \mathbf N)$. Now let $f \in F$ and $A \subseteq \mathbf R$ Borel. We have as $X=Y$ almost surely, that $$ \def\P{\mathbf P}\P(X \in A, f) = \P(Y \in A, f) = \P(Y \in A)\P(f) = \P(X\in A)\P(f)$$ Hence, $\sigma(X)$ and $F$ are independent.

Answer 2

Lemma: Two random variables $X$ and $Y$ are independent, if and only if $$\ \forall t_1,t_2\in\mathbb{R},\ \mathbb{E}\left[\exp\left(it_1X+it_2Y\right)\right]=\mathbb{E}\left[\exp\left(it_1X\right)\right]\mathbb{E}\left[\exp\left(it_2Y\right)\right]. $$

Proof of the original question:

• " $X$ is independent of $\mathcal{F}$ " is equivalent to " $\forall A\in\mathcal{F}$ , $X$ is independent of $I_A$ ".
• $X_n\overset{a.s.}{\rightarrow}X$ , hence $\exp\left(it_1X_n\right) \overset{a.s.}{\rightarrow} \exp\left(it_1X\right)$ and $\exp\left(it_1X_n+it_2I_A\right) \overset{a.s.}{\rightarrow} \exp\left(it_1X+it_2I_A\right)$.
• By dominated convergence theorem, we have that $\mathbb{E}\left[\exp\left(it_1X_n\right)\right] \rightarrow \mathbb{E}\left[\exp\left(it_1X\right)\right]$ and that $\mathbb{E}\left[\exp\left(it_1X_n+it_2I_A\right)\right] \rightarrow \mathbb{E}\left[\exp\left(it_1X+it_2I_A\right)\right]$ .
• For each $n$, $X_n$ is independent of $I_A$, hence $$\mathbb{E}\left[\exp\left(it_1X_n+it_2I_A\right)\right]=\mathbb{E}\left[\exp\left(it_1X_n\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right], $$
  then we have that
  $$\begin{align*} &\mathbb{E}\left[\exp\left(it_1X+it_2I_A\right)\right] = \lim_{n\rightarrow\infty}{\mathbb{E}\left[\exp\left(it_1X_n+it_2I_A\right)\right]} \\ &\qquad\qquad\qquad\qquad\quad =\lim_{n\rightarrow\infty}{\mathbb{E}\left[\exp\left(it_1X_n\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right]} \\ &\qquad\qquad\qquad\qquad\quad =\mathbb{E}\left[\exp\left(it_1X\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right]. \end{align*}$$
  It is concluded that $X$ is independent of $I_A$.