Convergence in probability of product and division of two random variables How can I prove the following:
Let $X_i$ and $Y_i$, $i = 1, \ldots, n$, $X$ and $Y$ be random variables defined on the probability space $(\Omega, \mathcal F, \mathbb P)$ and assume that $X_n$ converges in probability to $X$ and $Y_n$ to $Y$ (also in probability). Then:
If $Y_n \ne 0$ and $Y \ne 0$ almost surely, then $X_n/Y_n$ converges in probability to $X/Y$.
I tried the following:
Let $\epsilon > 0$. We need to show that
\begin{align*}
\mathbb P(|X_n/Y_n - X/Y| > \epsilon) \xrightarrow{n \to \infty} 0.
\end{align*}
Define $Z_n := 1/Y_n$ and $Z := 1/Y$. $Z_n$ and $Z$ are well defined, since $Y_n, Y \ne 0$ a.s. It is
\begin{align*}
\mathbb P(|Z_n-Z| > \epsilon) &= \mathbb P(|1/Y_n - 1/Y| > \epsilon) \\
&= \mathbb P(|Y-Y_n|/|Y_nY| > \epsilon) \\
&= \mathbb P(|Y_n-Y| > \epsilon |Y_n||Y|) \\
&\le \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|) \\
&= \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|, |Y_n-Y| \le |Y|/2) \\
&\quad + \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|, |Y_n-Y| > |Y|/2) \\
&\le \mathbb P(|Y_n-Y| > \epsilon Y^2/2) + \mathbb P(|Y_n-Y| > |Y|/2).
\end{align*} 
Now for any $A > 0$
\begin{align*}
\mathbb P(|Y_n-Y| > \epsilon Y^2/2) 
&= \mathbb P(|Y_n-Y| > \epsilon Y^2/2, |Y| \ge 1/A) + \mathbb P(|Y_n-Y| > \epsilon Y^2/2, |Y| < 1/A) \\
&\le \mathbb P\left(|Y_n-Y| > \frac{\epsilon}{2A^2}\right) + \mathbb P(|Y| < 1/A)
\xrightarrow{n \to \infty}  \mathbb P(|Y| < 1/A)
\end{align*}
and similarly
\begin{align*}
\mathbb P(|Y_n-Y| > |Y|/2) &= \mathbb P(|Y_n-Y| > |Y|/2, |Y| \ge 1/A) + \mathbb P(|Y_n-Y| > |Y|/2, |Y| < 1/A) \\
&\le \mathbb P\left(|Y_n-Y| > \frac{1}{2A}\right)  + \mathbb P(|Y| < 1/A)
\xrightarrow{n \to \infty} \mathbb P(|Y| < 1/A).
\end{align*}
Since we can choose $A$ arbitrarily large, we obtain
\begin{align*}
\mathbb P(|Z_n-Z| > \epsilon) \xrightarrow{n \to \infty} 0.
\end{align*}
The result follows since we know that $X_nZ_n$ converges to $XZ$ in probability.
Edit: Is it complete now? Is there maybe a shorter proof? Maybe my estimations are too long or there is shorter way to do it.
 A: Isn't this a consequence of the continuous mapping theorem?
Since Y is different to zero almost surely, the function g(x)=1/x satisfies the condition $P[Y\in C^c(g)]=0$, where $C^c(g)$ denotes the set of discontinuity points of $g$ (which in this case equals the set $\{0\}$). Hence we obtain $\frac{1}{Y_n}\to \frac{1}{Y}$ in probability.
Since $X_n\to X$ in probability and $Z_n\to Z$ in probability implies $X_nZ_n\to XZ$ in probability (this is proved in Resnick's "A Probability Path" using subsequences), the result follows taking $Z_n=\frac{1}{Y_n}$ and $Z=\frac{1}{Y}$.
A: The estimates look correct, but it seems not clear how to get the wanted convergence (we indeed have the probability that $|Y_n-Y|$ is greater than a positive number, but this one is random). 
However, the problem can be solved in this way: fix $\delta$ a small number and $A$ such that $\mathbb P\{|Y|\leqslant A\}\lt \delta$. Then we decompose the probabilities obtained in the last estimate of the OP computations over the set $\{|Y|\leqslant A^{-1}\}$ and its complement. 
For example, here is the proof that $\mathbb P\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\to 0$. We have $$\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}= (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\geqslant A\})\cup (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\lt A\})\\
\subset  \{|Y_n-Y|\geqslant \varepsilon A^2\}\cup\{Y\lt A\}.$$
