The convergence of a product I want to show the following theorem:
Suppose $(X_n)_{n \geq 0}$ converge in distribution to a random variable $X$. Another sequence $(Y_n)_{n \geq 0}$ of random variables converge in probability to $0$. Then $X_{n}Y_{n}$ converges in probability to $0$. My proof is as follows:
To show that $\lim_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|> \epsilon)=0$ for every $\epsilon>0$. Choose $\delta>0$, we have
\begin{align*}
    \mathbb {P}(|X_{n}Y_{n}|> \epsilon)=\mathbb {P}(|X_{n}Y_{n}|> \epsilon,|Y_{n}|\leq \delta)+\mathbb {P}(|X_{n}Y_{n}|> \epsilon,|Y_{n}|>\delta).
\end{align*}
Given that $|X_{n}Y_{n}|> \epsilon$ and $|Y_{n}|\leq \delta$, we have $|X_{n}|>\epsilon/\delta$. Since $\mathbb {P}(|X_{n}Y_{n}|> \epsilon,|Y_{n}|>\delta)\leq \mathbb {P}(|Y_{n}|>\delta)$, we have
\begin{align*}
    \mathbb {P}(|X_{n}Y_{n}|> \epsilon)\leq \mathbb {P}\left(|X_{n}|> \frac {\epsilon}{\delta}\right)+\mathbb {P}(|Y_{n}|>\delta).
\end{align*}
Since we know that $\lim_{n\to \infty}\mathbb {P}(|Y_{n}|>\delta)=0$, we have
\begin{align*}
    \limsup_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|>\epsilon)\leq \limsup_{n\to \infty}\mathbb {P}\left(|X_{n}|>\frac {\epsilon}{\delta}\right)=\mathbb {P}\left(|X|>\frac {\epsilon}{\delta}\right).
\end{align*}
By taking $\delta \to 0^{+}$, we have
\begin{align*}
    \lim_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|>\epsilon)=0.
\end{align*}
Do I need to assume that $F_X$ has at most countable number of discontinuities?
 A: $\lim \sup P(|X_n| >\frac {\epsilon} {\delta}) \leq \lim \sup P(|X_n| >\frac {\epsilon'} {\delta})$ whenever $0 <\epsilon' <\epsilon$. We can choose $\epsilon' $ such that $\frac {\epsilon'} {\delta}$ is a continuity point. (Can you see why? Prove this by contradiction). Now let $\delta \to 0$ as before. 
[If $\epsilon'$ does not exist as stated then there are uncountably many discontinuity points, which is a contradiction]. 
A: In order to solve the technical difficulty of the continuity point, let $\varepsilon$ be fixed pick a sequence 
$\left(\delta_k\right)_{k\geqslant 1}$ converging to $0$ such that for all $k$, $\varepsilon/\delta_k$ is a continuity point of the cumulative distribution function of $X$ and $-X$. This is possible: we can choose $\delta_k\in (0,1/k)$, because the number of discontinuities of the cumulative distribution function of a random variable (here $X/\varepsilon$) is at most countable. 
You proved that for all fixed $k$, 
$$
 \limsup_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|>\epsilon)\leq \limsup_{n\to \infty}\mathbb {P}\left(|X_{n}|>\frac {\epsilon}{\delta_k}\right)
$$
and with the previous restriction on $\delta_k$, we get for all $k$. 
$$
 \limsup_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|>\epsilon)\leq  \mathbb {P}\left(|X |>\frac {\epsilon}{\delta_k}\right).
$$
