Almost sure convergence of product of two random variable Suppose X1, X2 ... converges to a with probability 1 , and Y1, Y2 ... converges to b with probability 1. Show that XnYn converges to ab with probability 1.
So I am learning convergence of random variables now and I haven't learnt anything about probability space. I don't know where to start when proving almost sure convergence. One method I'm familiar with is to show the expectation of the sum of a sequence of random variables is finite, then this sequence will eventually converge to 0. However in general, such as this problem, I don't know how to deal with the definition of almost sure convergence. Since in the definition, the  limit operator is inside the probability operator and I don't think I can use the ε - δ technique to translate the definition, which is useful for convergence in probability though.
 A: The definition of the limits extends nicely to the probabilities.
You just have to keep in mind that a statement like $X=x$ is a event, that is a set $\lbrace \omega \in\Omega \mid X(\omega)=x\rbrace$. The definition of almost sure convergence is 
$$\mathbb{P}[\lim_{n\rightarrow \infty} X_n = x]=1$$
Let's first write the definition of $\lim_{n\rightarrow \infty} x_n=x$ as $\forall \epsilon \exists n_0$ such that $\forall n\geq n_0, \lvert x_n - x \rvert < \epsilon$. If we look at a similar statement, $\lvert X_n - x \rvert < \epsilon$ is now a set. When dealing with sets, a $\forall$ statement correspond to an intersection and a $\exists$ correspond to a union. Let's get some intuition about that, if you have a $\forall$ statement, then whenever it fails, the statement is wrong. For sets this corresponds to an intersection since as long as one element is not in one of the sets, then it is not in the intersection. The reverse happens for $\exists$ statements.
$$
\cup_{n_0>0} \cap_{n\geq n_0} \lbrace \omega \in  \Omega \mid \lvert X_n(\omega) - x \rvert < \epsilon \rbrace
$$
Hope this makes a bit clearer. Note that since $\epsilon\in \mathbb{R}$, we cannot take the union, but we can actually take out the $\forall \epsilon$ to get
\begin{align*}
&\mathbb{P}[\lim_{n\rightarrow \infty} X_n = x]=1\\
\Leftrightarrow & \forall \epsilon , \mathbb{P}[\cup_{n_0>0} \cap_{n\geq n_0} \lbrace \omega \in  \Omega \mid \lvert X_n(\omega) - x \rvert < \epsilon \rbrace]=1
\end{align*}
Regarding your question, given $\epsilon>0$, suppose that for some $n$, $\lvert X_n-a\rvert < \epsilon$ and $\lvert Y_n-b\rvert < \epsilon$, observe that
\begin{align*}
\epsilon^2 &> \lvert (X_n-a)(Y_n-b) \rvert\\
&=\lvert X_nY_n-a Y_n - b X_n + a b \rvert\\
&=\lvert X_nY_n-a (Y_n-b) - b (X_n-a) - a b \rvert\\
&\geq \lvert X_nY_n - a b \rvert - \lvert a (Y_n-b)\rvert  - \lvert b (X_n-a)\rvert\\
&> \lvert X_nY_n - a b \rvert - (\lvert a \rvert + \lvert b \rvert) \epsilon
\end{align*}
Which shows the statement since then $\epsilon^2+(\lvert a \rvert + \lvert b \rvert) \epsilon > \lvert X_nY_n - a b \rvert$, on set notation it means that 
\begin{align*}
\lbrace \lvert X_n-a\rvert < \epsilon \rbrace\cap \lbrace \lvert Y_n-b\rvert < \epsilon \rbrace \subset \lbrace \lvert X_nY_n - a b \rvert < \epsilon^2 +(\lvert a \rvert + \lvert b \rvert) \epsilon\rbrace
\end{align*}
Because for any $\epsilon'>0$, we can always find $\epsilon>0$ such that $\epsilon^2 +(\lvert a \rvert + \lvert b \rvert) \epsilon<\epsilon'$, we can combine that with the first comment of your question (Michael's), you can get to your desired result.
