$X_n \to X, Y_n \to c$ in distribution implies $X_n Y_n \to Xc$ in distribution

Question

I am trying to prove $$X_n \xrightarrow{d} X, Y_n \xrightarrow{d} a \implies Y_n X_n \xrightarrow{d} aX$$ where $a$ is a constant.
What I tried:
Let $g:\mathbb R\to \mathbb R$ an arbitrary uniformly continuous, bounded function. It suffices to show $\mathbb E[g(Y_n X_n)] \to \mathbb E[g(aX)]$. We have $$\left \lvert \int g(Y_nX_n) - g(aX) \,dP \right \rvert \leq \left \lvert \int g(Y_n X_n) - g(aX_n) \, dP \right \rvert +\left \lvert \int g(aX_n) - g(aX) \, dP \right \rvert,$$ where the right summand goes to $0$ by assumption as $a$ is constant. Now I want to use uniform continuity of $g$ to estimate the left summand: Choose $\delta > 0 $ such that $$\left \lvert g(Y_n X_n) - g(aX_n) \right \rvert < \epsilon,$$ whenever $|Y_n X_n - aX_n| < \delta.$ Since convergence in distribution to a constant implies convergence in probability, we have can control $P(|Y_n - a | > \delta)$. But is it possible to get a bound on $|X_n|$? I assume that probability in distribution does not imply some form of boundedness... Any help is appreciated.

Answer 1

First, you can choose $g$ to be Lipschitz bounded, with $C$ its Lipschitz constant. It follows that $\forall \epsilon >0, A > 0$: $$|\mathbb{E}g(X_n Y_n) - \mathbb{E}g(X_n a)| \leq \mathbb{E}|g(X_n Y_n) - g(X_n a)|1_{|X_n Y_n - X_n a| > \epsilon} + \mathbb{E}|g(X_n Y_n) - g(X_n a)|1_{|X_n Y_n - X_n a| \leq \epsilon} \leq C \epsilon + ||g||_\infty \mathbb{P} (|X_n Y_n - X_n a | > \epsilon) \leq C \epsilon + \mathbb{P} (|X_n Y_n - X_n a | > \epsilon, |X_n|> A) + \mathbb{P} (|X_n Y_n - X_n a | > \epsilon, |X_n| \leq A) \leq C \epsilon + \mathbb{P} (|Y_n - a | > \epsilon / A) + \mathbb{P} (|X_n| \geq A)$$

Now, $\lim_{n \rightarrow \infty} \mathbb{P} (|Y_n - a | > \epsilon / A) = 0$ so $\limsup{n \rightarrow \infty} |\mathbb{E}g(X_n Y_n) - \mathbb{E}g(X_n a)| \leq C \epsilon + \mathbb{P} (|X| \geq A)$. I used here that fact that $X_n$ goes in law to $X$ and that $[A, \infty[$ is closed (see the Portemanteau lemma). You can now let $\epsilon$ goes to zero and $A$ goes to $\infty$.