Let $X_1, X_2, \ldots$ be a sequence of random variables on a probability space $(\Omega, \mathcal{F}, P)$, and let $\mathcal{G}$ be a $\sigma$-subalgebra of $\mathcal{F}$. Assume $X_n$ is independent of $\mathcal{G}$ for each $n$, and that $\lim_{n\to\infty} X_n = X$ almost surely. Prove that $X$ is independent of $\mathcal{G}$.
This result makes sense to me intuitively, but I'm not sure how to prove the result. For some reason, this problem is in the "Characteristic Functions" portion of my probability textbook, so maybe there's some clever way to solve this problem with characteristic functions (I would be satisfied to see any solution though)?
I'm not so sure where to start
Not true unless $\mathcal G$ is complete under $P$ or $X_n \to X$ at every point.
If $\mathcal G$ is not complete there exists $A \in \mathcal G$ and a set $B \in \mathcal F$ such that $P(A\Delta B)=0$ and $B \notin \mathcal G$. Take $X_n=I_A$ for all $n$ and $X=I_B$ to get a counter-example.
If $X_n(\omega) \to X(\omega)$ for every $\omega$ then this result is immediate from the fact that pointwise limit of a sequence of measurable functions on $(\Omega,\mathcal G)$ is measurable.