Types of convergence $X_n \to X$ under which $X$ is independent of $Y$ I can't seem to find this kind of result anywhere so let me ask here.
Let $\{X_n\}_{n \geq 0}$ random variables be independent of $Y$ for all $n$. Under which types of convergence $X_n \to X$ is $X$ independent of $Y$? And why?
 A: Suppose that $\{X_n\}$ is independent of $Y$ in the sense that for any $n$, the random vector $(X_1, \dots, X_n)$ is independent of $Y$.  Then it follows that the $\sigma$-field $\sigma(X_1, X_2, \dots)$ generated by all $X_n$ is also independent of $Y$.  (Our assumption gives us that for each $n$, the $\sigma$-field $\sigma(X_1, \dots, X_n)$ is independent of $Y$.  A monotone class or $\pi$-$\lambda$ argument lets us conclude that $\sigma\left(\bigcup_n \sigma(X_1, \dots, X_n)\right) = \sigma(X_1, X_2, \dots)$ is also independent of $Y$.)
If $X_n \to X$ surely, then $X$ is itself $\sigma(X_1, X_2, \dots)$-measurable (a limit of measurable functions is measurable).  Hence $X$ is independent of $Y$.  The same holds if $X_n \to X$ almost surely.
If $X_n \to X$ in probability, we may find a subsequence $X_{n_k} \to X$ almost surely, so by the previous result $X$ is again independent of $Y$.  This also covers the case when $X_n \to X$ in $L^p$ for some $p \ge 1$.
Weak convergence (aka convergence in distribution) is not sufficient, since it only determines the limiting random variable up to equality in distribution.  For example, let $X_1, X_2, \dots, Y$ be iid with a non-constant distribution.  It follows from the definition that $X_n \to Y$ weakly, but $Y$ is not independent of itself.
Edit: In his comment below, Did points out the following alternate proof, which also covers the case where we merely assume that for each $n$, $X_n$ is independent of $Y$.  Suppose $u,v$ are bounded continuous functions; by independence, $E[u(X_n) v(Y)] = E[u(X_n)] E[v(y)]$.  Using the dominated convergence theorem, we pass to the limit to obtain $E[u(X) v(Y)] = E[u(X)] E[v(Y)]$.  By a monotone class argument, this also holds for $u,v$ any bounded measurable functions, and hence $X$ and $Y$ are independent.  As before, this argument works directly when $X_n \to X$ almost surely, and by passing to a subsequence it also works when $X_n \to X$ in probability or in $L^p$.
A: Just to complete Nate's answer, note that the convergence of $\{X_n\}$ itself can depend on $Y$, even if every $X_n$ is individually independent of $Y$.
For a trivial example, let $X_0$ and $Y$ be independent and uniformly distributed over $\{-1, 1\}$, and let $X_n = YX_{n-1}$ for $n > 0$.  It is easy to see that every $X_n$ is individually independent of $Y$.
Clearly, if $Y = 1$, then $X_n = X_{n-1} = X_0$ for all $n > 0$, and the sequence converges in any sense you care to name; whereas, if $Y = -1$, then $X_n = -X_{n-1}$ and it does not.
