Convergence in probability of the product of two random variables Suppose $\{X_n\}$ and $\{Y_n\}$ converge in probability to $X$ and $Y$, respectively.
Will $X_n Y_n$ converge in probability to $X Y$?
I know the answer is yes.
If we treat $(X_n,Y_n)$ as a random vector, and it converges in probability to $(X,Y)$ by the assumption. Then $g(x,y) = xy$ is a continuous function and according to the continuous mapping theorem, $g(X_n,Y_n)$ converges in probability to $g(X,Y)$.
My question is how to go from the definition without using the continuous mapping theorem.
My attempt is as follows.
$$P(|X_nY_n-XY|>\epsilon)=P(|X_nY_n-X_nY+X_nY-XY|>\epsilon)$$
$$\leq P(|X_n(Y_n-Y)|+|Y(X_n-X)|>\epsilon)$$
It seems tempting to conclude that the last term goes to zero as $n$ goes to infinity.
But I am not sure about it.
Am I right or did I miss something?
 A: This is pretty straightforward if you use that

$X_n$ tends to $X$ in probability if, and only if, every subsequence of $X_n$ has a sub(sub)sequence that tends to $X$ a.s.

This lemma follows from:
Fact 1. If $X_n$ tends to $X$ a.s., then $X_n$ tends to $X$ in probability.
Fact 2. If $X_n$ tends to $X$ in probability, it has a subsequence that tends to $X$ a.s.
Fact 3. Let $(a_n)$ be a sequence of real numbers. Then $(a_n)$ converges to $a \in \Bbb R$ if, and only if, every subsequence of $(a_n)$ has a sub(sub)sequence that tends to $a$.
Application
Let $(X_{\phi(n)}Y_{\phi(n)})$ be a subsequence of $(X_nY_n)$. We need to show that it admits a subsequence converging to $XY$ a.s. Since $X_n$ tends to $X$ in probability, there exists $\psi$ such that $X_{\phi(\psi(n))}$ tends to $X$ a.s. Since $Y_n$ tends to $Y$ in probability, there exists $\chi$ such that $Y_{\phi(\psi(\chi(n)))}$ tends to $Y$ a.s. Now, remark that $X_{\phi(\psi(\chi(n)))}Y_{\phi(\psi(\chi(n)))}$ tends to $XY$ a.s.
A: For every $\varepsilon\gt0$ and $u\geqslant0$, let $\alpha_{u,\varepsilon}=\varepsilon(u+2\varepsilon)$. Then
$$
[|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon}]\subseteq[|X_n-X|\geqslant\varepsilon]\cup[|Y_n-Y|\geqslant\varepsilon]\cup[|X|\geqslant u]\cup[|Y|\geqslant u].
$$
(Proof: If $|x_n-x|\lt\varepsilon$, $|y_n-y|\lt\varepsilon$, $|x|\lt u$ and $|y|\lt u$, then $|x_ny_n-xy|\lt\varepsilon(u+2\varepsilon)$.)
Hence,
$$
\mathbb P(|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon})\leqslant\mathbb P(|X_n-X|\geqslant\varepsilon)+\mathbb P(|Y_n-Y|\geqslant\varepsilon)+\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u).
$$
Consider the limit $n\to\infty$. One gets
$$
\limsup_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon})\leqslant\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u).
$$
For every $\eta\gt0$ and $u\gt0$, there exists $\varepsilon$ such that $\eta\geqslant\alpha_{u,\varepsilon}$, thus
$$
\limsup_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\eta)\leqslant\inf\limits_{u\gt0}\left(\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u)\right).
$$
The infimum on the RHS is zero hence, for every $\eta\gt0$,
$$
\lim_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\eta)=0.
$$
A: 
It seems tempting to conclude that the last term goes to zero as n goes to infinity. But I am not sure about it. Am I right or did I miss something?

You're right, this can be done directly and we only need a little bit more work to control the right term. The following inclusion of events is easy to check:
$$
\{|X_n(Y_n-Y)|+|Y(X_n-X)|>\epsilon\}\subset \{|X_n|\cdot|Y_n-Y|>\epsilon/2\}\cup\{|Y|\cdot|X_n-X|>\epsilon/2\}.
$$
Now, for any $A>0$,
$$
\{|X_n|\cdot|Y_n-Y|>\epsilon/2\}\subset \{|X_n-X|>1\}\cup\{|X+1|> A\}\cup\{|Y_n-Y|>\epsilon/2(A+1)\}
$$
Hence, using the convegence in probability of $X_n$ to $X$ and of $Y_n$ to $Y$, we deduce that, for any $A>0$,
$$
\limsup_{n\rightarrow\infty}\mathbb{P}(|X_n|\cdot|Y_n-Y|>\epsilon/2)\leq \mathbb{P}(|X+1|> A)\xrightarrow[A\rightarrow\infty]{}0.
$$
Similarily (in fact easier), $\mathbb{P}(|Y|\cdot|X_n-X|>\epsilon/2)$ goes to $0$ when $n\rightarrow\infty$. This concludes your proof!
A: First,
\begin{align}
|X_nY_n-XY| &\le |(X_n-X)(Y_n-Y)|+|(X_n-X)Y|+|(Y_n-Y)X| \\
&\le \frac{1}{2}(X_n-X)^2+\frac{1}{2}(Y_n-Y)^2+|(X_n-X)Y|+|(Y_n-Y)X|.
\end{align}
Then for any $\epsilon>0$ and $K>0$
\begin{align}
P\{|X_nY_n-XY|>\epsilon\} &\le P\{|X_n-X|>\sqrt{\epsilon/2}\}+ P\{|Y_n-Y|>\sqrt{\epsilon/2}\} \\
&+P\{|X_n-X|>\epsilon/(4K)\}+P\{|Y_n-Y|>\epsilon/(4K)\} \\
&+P\{|X|>K\}+P\{|Y|>K\}; \\
\\
\because \{|(X_n-X)Y|>\epsilon/4\}&\subset\{|Y|>K\}\cup \{|X_n-X|>\epsilon/(4K)\},\\
\{|(Y_n-Y)X|>\epsilon/4\}&\subset\{|X|>K\}\cup \{|Y_n-Y|>\epsilon/(4K)\}.
\end{align}
For any $\nu>0$ we can find $K$ s.t. $P\{|X|>K\}<\nu$ and $P\{|Y|>K\}<\nu$ so that
$$\limsup_{n\to\infty}P\{|X_nY_n-XY|>\epsilon\}\le 2\nu.$$
Now, send $\nu\downarrow 0$.
