Let $\{X_n\}$ and $\{Y_n\}$ be two sequences of random variables such that $X_n$ converges to $X$ in probability and $Y_n$ converges to $Y$ in probability. Show that $X_n Y_n$ converges to $XY$ in probability.
Hence, we have to show that: $$ X_n\stackrel{P}\rightarrow X\text{, }Y_n\stackrel{P}\rightarrow Y \Rightarrow X_nY_n\stackrel{P}\rightarrow XY $$
I was thinking about some way to 'rewrite' $X_nY_n$. In particular, $$X_nY_n=X_n(Y_n-X_n)+X_n^2$$ At this point, I was thinking that, for a sufficiently large $M\in\mathbb{N}$, for $n\geq M$, since $\lim(a\cdot b)=\lim(a)\cdot\lim(b)$ and $\lim(a+b)=\lim(a)+\lim(b)$, $$\lim\limits_{n}(X_nY_n)=\lim(X_n)\cdot\lim(Y_n)-\lim(X_n)\cdot\lim(X_n)+\lim(X_n^2)=$$ $$=\lim(X_n)\cdot\lim(Y_n)-\lim(X_n)\cdot\lim(X_n)+\lim(X_n)\lim(X_n)=$$ $$=X\cdot Y -X^2 +X^2=XY$$ However, I am pretty sure this is not the right way for the following reason:
convergence in probability of $X_nY_n$ to $XY$ requires that $\lim\limits_{n}\mathbb{P}\left(|X_nY_n-XY|>\varepsilon\right)=0$ and I think that my reasoning - should it be correct - is not sufficient.
How could I proceed to solve such an exercise alternatively?
We use the following equivalence:
Now, we show $X_n Y_n \stackrel{\mathbb{P}}\to XY$. Let $(X_{n_k} Y_{n_k})_k$ be a subsequence of $(X_nY_n)_n$. Then $(X_{n_k})$ has a further subsequence $(X_{n_{k_l}})_l$ that converges almost surely to $X$ and $(Y_{n_{k_l}})_l$ has a further subsequence $(Y_{n_{k_{l_j}}})_j$ that converges almost surely to $Y$. Then since also $(X_{n_{k_{l_j}}})_j$ converges to $X$ almost surely, we can deduce that $X_{n_{k_{l_j}}}Y_{n_{k_{l_j}}} \to XY$ almost surely (refer to $X_n\stackrel{a.s.}\rightarrow X, Y_n \stackrel{a.s.}{\rightarrow} Y \implies X_n Y_n \stackrel{a.s.}\rightarrow XY$).
We can conclude, because we have found a further subsequence of $(X_{n_k}Y_{n_k})_k$ that converges almost surely.
I have to admit that this looks ugly but the idea is simple.