@martini's answer is correct. But he(or She) fails to mention something which may confuse you. I'm going to give more explanations on his(or her) answer.
What we have by Borel Cantelli lemma is
$P(\cap_{n=1}\cup_{k\geq n}\{X_n\neq Y_n\}) = 0$
i.e.
$P(\cup_{n=1}\cap_{k\geq n}\{X_n = Y_n\}) = 1$
By using the VERY important fact that
$\big\{\omega:\cup_{n=1}\cap_{k\geq n}\{X_n(\omega) = Y_n(\omega)\}\big\} \subseteq\big\{\omega: \lim X_n(\omega) =\lim Y_n(\omega) \big\}$
the above relation is very important, but most of the text books fail to mention. You can prove it yourself and it's not hard. Thus we have
$P( \lim X_n =\lim Y_n ) = 1,\quad \text{i.e.}\quad \lim X_n =\lim Y_n\quad \text{a.s.}$
By using another lemma that, for real numbers $a_n\in R$, if
$\lim a_n = a, \text{ then }\lim_{n\to\infty} \frac{\sum_{j=1}^na_j}{n} = a$
we thus have
$\big\{\omega: \lim X_n(\omega) =\lim Y_n(\omega) \big\}\subseteq\big\{\omega: \lim \frac{\sum_{k=1}^nX_j(\omega)}{n} =\lim \frac{\sum_{k=1}^nY_j(\omega)}{n} \big\}$
Finnaly
$P\big(\big\{\omega: \lim \frac{\sum_{k=1}^nX_j(\omega)}{n} =\lim \frac{\sum_{k=1}^nY_j(\omega)}{n} \big\}\big)=1$
Then we can say it's a complete prrof.