I am trying to show the following- If $X_n$ converges in distribution to $X$ and $Y_n$ converges in probability to 0, then show that $X_nY_n$ converges in probability to 0.
Attempt- $$P(|X_nY_n|>\epsilon) = P(|X_nY_n|>\epsilon,|Y_n|>\eta) + P(|X_nY_n|>\epsilon,|Y_n|\leq\eta) \leq P(|Y_n|>\eta) +P(|X_n|>\epsilon/\eta)$$
Now the term on the left goes to 0 but what do I o with the remaining part, can someone please provide the solution.
Let $\varepsilon > 0$ and $\eta_k \searrow 0$, such that $\varepsilon / \eta_k$ is a continuity point of $F(x) : = \Bbb P (|X| \leq x)$, (this is possible since $F$ has at most countable discontinuities) $$\Bbb P (|X_n||Y_n| > \varepsilon) = \Bbb P (|X_n||Y_n| > \varepsilon , |Y_n| > \eta_k) + \Bbb P (|X_n||Y_n| > \varepsilon , |Y_n| \leq \eta_k)\\ \leq \Bbb P (|Y_n| > \eta_k) + \Bbb P (|X_n| > \varepsilon / \eta_k) = \Bbb P (|Y_n| > \eta_k) + 1 - F_n(\varepsilon / \eta_k),$$ where $F_n(x) := \Bbb P(|X_n| \leq x)$. Hence $$\limsup_{n\to\infty} \Bbb P (|X_n||Y_n| > \varepsilon) \leq 1 -F(\varepsilon / \eta_k) \searrow 0,$$ as $k\to\infty$.
