Product of a sequence converging in distribution and a sequence converging in probability converges in distribution

Question

I'm working on the following problem and would appreciate any hints or solutions

Suppose $X_n\to X$ in distribution and $Y_n\to 1$ in probability. Show that $X_nY_n\to X$ in distribution.

Here are my thoughts: $Y_n$ converging in probability to 1 means $P(1-\epsilon<Y_n<1+\epsilon)\to 1$ as $n\to \infty$. I would like to choose $\epsilon$ less than 1 so that we don't have to worry about multiplying any negative numbers to inequalities and then I want to use that to show that $P(X_nY_n\leq x) \to P(X\leq x)$. I'm having trouble putting these two ideas together though, by somehow multiplying by $Y_n$. I suspect this could also involve taking conditional probabilities.

Source: Spring 1997

Answer 1

Hint: \begin{align} &P(X_n(1+\epsilon) \le x)\ P(|Y_n - 1| < \epsilon) \\ &\le P(X_n Y_n \le x) \\ &\le P(X_n(1-\epsilon) \le x)\ P(|Y_n - 1| < \epsilon) + P(|Y_n - 1| > \epsilon) \end{align}