Product of random variables convergence in probability I want to prove that if $X_n\overset{P}\to X$ and $Y_n\overset{P}\to Y$, then
$$X_n+Y_n\overset{P}\to X+Y,$$
$$X_nY_n\overset{P}\to XY$$
Proof for the first one I simply use triangle inequality
$$
\mathbf{P}\left(\omega \mid |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|<\epsilon\right) \leqslant \mathbf{P}\left(\omega \mid |(X_n(\omega)-X(\omega))|+|Y_n(\omega)-Y(\omega)|<\epsilon\right)\leqslant$$
$$\leqslant \mathbf{P}\left(\omega | |X_n(\omega)-X(\omega)|<\frac{\epsilon}2\right)+\mathbf{P}\left(\omega | |Y_n(\omega)-Y(\omega)|<\frac{\epsilon}2\right)\underset{n}\to 0$$
I could use some ideas to prove convergence for the product.
 A: Obviously,
$$|X_n Y_n-XY| \leq |Y| \cdot |X_n-X|+ |X_n| \cdot |Y_n-Y|.$$
By the triangle inequality,
$$\mathbb{P}(|X_n Y_n-XY| \geq \epsilon) \leq \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2) + \mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2).$$
We estimate the terms separately. For the first one, note that
$$\mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2)\leq \mathbb{P}(|X_n-X| \geq \epsilon/(2M)) + \mathbb{P}(|Y| \geq M)$$
for any $M>0$. Letting $n \to \infty$, we find
$$\limsup_{n \to \infty} \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2) \leq \mathbb{P}(|Y| \geq M).$$
Since this holds for any $M$, we can let $M \to \infty$ and, e.g. by the dominated convergence theorem, it follows that the limit $\lim_{n \to \infty} \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2)$ exists and equals $0$.
The second term is a little bit more difficult, but the idea is the same. Again, we estimate
$$\mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) \leq \mathbb{P}(|Y_n-Y| \geq \epsilon/(2M)) + \mathbb{P}(|X_n| \geq M). \tag{1}$$
Now, by the triangle inequality,
$$\mathbb{P}(|X_n| \geq M) \leq \mathbb{P}(|X_n-X| \geq M/2) + \mathbb{P}(|X| \geq M/2). \tag{2}$$
Plugging $(2)$ into $(1)$ yields
$$\mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) \leq \mathbb{P}(|Y_n-Y| \geq \epsilon/(2M))+ \mathbb{P}(|X_n-X| \geq M/2) + \mathbb{P}(|X| \geq M/2).$$
The first two terms on the right-hand side converge to $0$ as $n \to \infty$ because $X_n \to X$ and $Y_n \to Y$ in probability. After letting $n \to \infty$ we also let $M \to \infty$ and this proves
$$\lim_{n \to \infty} \mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2) =0.$$
