Independence of a sequence converging in probability implies independence of the limit

Question

Let ${X_1,X_2,\dots}$ be a sequence of scalar random variables converging in probability to another random variable ${X}$. Suppose that there is a random variable ${Y}$ which is independent of ${X_i}$ for each individual ${i}$. Show that ${Y}$ is also independent of ${X}$.

By splitting into real and imaginary parts, we may assume without loss of generality that the $X_i$ and $X$ are real-valued. As

$$ \displaystyle {\bf P}(X \leq t \wedge Y \in S) = {\bf P}((|X-X_i|>\varepsilon \wedge X \leq t)) \wedge Y \in S) + {\bf P}((|X-X_i| \leq \varepsilon \wedge X \leq t)) \wedge Y \in S) $$

and the first term on the RHS tends to zero for large $i$, my crude first impression was to approximate the second term on the RHS by ${\bf P}(X_i \leq t \wedge Y \in S) = {\bf P}(X_i \leq t){\bf P}(Y \in S)$, and then the claim follows if one can show that ${\bf P}(X_i \leq t)$ approximates ${\bf P}(X \leq t)$.

Yet I struggles a bit to ground all these on a rigorous base, some helps or even a completely different approach will be appreciated.

Answer 1

To establish the desired independence, it suffices to show that $$ E[f(X)g(Y)] = E[f(X)]\cdot E[g(X)] $$ for all bounded continuous functions $f$ and $g$ mapping $\Bbb R$ to itself. Because $(X_n)$ converges in probability to $X$, there is a subsequence $1\le n(1)<n(2)<\cdots$ such that $\lim_kX_{n(k)}=X$ a.s. Then, by dominated convergence $$ E[f(X)g(Y)]=\lim_k E[f(X_{n(k)})g(Y)]=\lim_k E[f(X_{n(k)})]\cdot E[g(Y)]=E[f(X)]\cdot E[g(Y)]. $$ The assumed independence was used for the second equality above.

Answer 2

Since $X_i\overset{p}{\to}X$, there is a subsequence $(X_{i_n})_{n=1}^\infty$ of $(X_i)_{i=1}^\infty$ which converges almost surely to $X$. We can relabel to assume that the original sequence $(X_i)_{i=1}^\infty$ converges a.s. to $X$. Then for any $t\in\mathbb{R}$, $$(X\leqslant t)=\bigcap_{l=1}^\infty \bigcup_{m=1}^\infty \bigcap_{n=m}^\infty (X_n\leqslant t+1/l).$$ Since each $X_i$ is independent of $Y$, every member of the sigma algebra $\sigma(X_1,X_2,\ldots)$ generated by $X_1,X_2,\ldots$ is also independent of $Y$. We just showed that $(X\leqslant t)\in \sigma(X_1,X_2,\ldots)$, and is therefore independent of $Y$.