Need Critique on my solution. Convergence in probability of a product of sequences of random variables Came across this problem in my self education. Found 4 solution here, but none looked simple enough to me... So I cooked up one of my own. Now it looks too simple :) Am I missing something?
I would really appreciate any constructive critique on my solution
 A: Let me summarize the issues we already discussed (in the comments):


*

*You cannot apply  the theorem of boundedness of convergent sequences since this would require that $X_n(\omega)$ converges for each $\omega \in \Omega$ (i.e. almost sure convergence) - but here only convergence in probability is assumed. In fact, it is not difficult to construct a sequence $(X_n)_{n \in \mathbb{N}}$ which converges in probability but is unbounded.

*At the end of the proof you use $$\mathbb{P} \left( |Y_n-Y| > \frac{\epsilon}{2M} \right) \to 0 \quad \text{as} \, \, n \to \infty$$ for a random variable $M$. Mind that convergence in probability does only imply $$\mathbb{P} \left( |Y_n-Y| > \frac{2\epsilon}{c} \right) \to 0$$ for some constant $c>0$ (by "constant" I mean that it does not depend on $\omega$).


Here is a sketch how to fix your proof:

Using a very similar argumentation as in your question, we can show that
$$\mathbb{P}(|X_n Y_n-XY| \geq \epsilon) \leq \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) + \mathbb{P} \left( |X| \cdot |Y_n-Y| > \frac{\epsilon}{2} \right) =: I_1+I_2.$$
We estimate the terms separately. For $I_1$ note that
$$\begin{align*} I_1 &\leq \mathbb{P}\left(|Y_n| <R, |Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) + \mathbb{P} \left( |Y_n| \geq R, |Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) \\ &\leq \mathbb{P} \left( |X_n-X| > \frac{\epsilon}{2R} \right) + \mathbb{P}(|Y_n| \geq R) \tag{1}. \end{align*}$$
Now
$$\mathbb{P}(|Y_n| \geq R) \leq \mathbb{P}\left(|Y_n-Y| \geq \frac{R}{2} \right) + \mathbb{P} \left( |Y| \geq \frac{R}{2} \right). \tag{2}$$
If we plug $(2)$ into $(1)$ and let $n \to \infty$, we get
$$\limsup_{n \to \infty} \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right)\leq \mathbb{P} \left( |Y| \geq \frac{R}{2} \right).$$
Finally, we can let $R \to \infty$ and conclude
$$\lim_{n \to \infty} \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) = 0.$$
A similar (even easier) argumentation applies to $I_2$. Consequently,
$$\lim_{n \to \infty} \mathbb{P}(|X_n Y_n-XY|  \geq \epsilon)=0.$$
A: Thanks to the great help from saz, I made an attempt to complete the proof(but found out later that the very same way was already presented here)
Never the less, Many thanks to saz!!!
A: I don't like these sorts of proofs that unnecessarily dive into a sea of $\epsilon$'s. Consider the following alternative characterization of convergence in probability.
Theorem. $X_n \to X$ in probability if and only if for every subsequence of the $X_n$ there is a further subsequence that converges to $X$ almost surely.
See, e.g. Williams: Probability with Martingales exercise EA13.1.
Using this theorem, let's prove your result. Suppose $X_n \to X, Y_n \to Y$ in probability. Let a subsequence $X_{n_k}Y_{n_k}$ of $X_nY_n$ be given. By the theorem take a subsequence and then another subsequence so that we have a subsequence on which $X_{n_{k_\ell}},Y_{n_{k_\ell}}$ both converge almost surely. Then of course $X_{n_{k_\ell}}Y_{n_{k_\ell}}$ converges almost surely to $XY$, so the theorem implies $X_nY_n \to XY$ in probability.
I encourage you to try and prove this theorem, it is not very difficult (though it will require some $\epsilon$'s which are, in my opinion, better spent on the theorem than your result.)
