# Show that $Y_n$ converges to Y almost surely

Let $Y_1$, $Y_2$,... be a sequence of random variables on a probability space ($\Omega$ , F , $\mu$). Assume that there is a random variable Y on the same probability space such that for any $\epsilon$>0, $$\sum_{n=1}^\infty P\{\lvert Y_n-Y \lvert > \epsilon \} < \infty.$$ Show that $Y_n$ converges to Y almost surely.

What I believe to be the definition of almost sure convergence is,

$$P(w: \lim_{n \rightarrow \infty}Y_n = Y) = P( \bigcup_{\epsilon>0 \ rationals} \bigcap_{m=1}^{\infty} \bigcup_{n > m}^{\infty} \ \{ |Y_n - Y| < \epsilon \}) = 1$$

Like the answer below suggested. We the apply the Borel-Cantelli Lemma which states

$$\ If \ \sum_{n=1}^{\infty} P(A_n)< \infty, \ then \ P(A_n \ i.o) = P( \limsup_{n \rightarrow \infty}\ A_n) = P(\bigcap_{m=1}^{\infty} \bigcup_{n = m}^{\infty}A_n)=0.$$ From our assumptions, since we know for any $\epsilon>0$, $\sum_{n=1}^\infty P\{\lvert Y_n-Y \lvert > \epsilon \} < \infty$. If we define $A_n= {|Y_n -Y|>\epsilon}$, we can say $$P(\bigcup_{\epsilon>0 \ rationals \ } \bigcap_{m=1}^{\infty} \bigcup_{n = m}^{\infty} \{|Y_n-Y|> \epsilon \})=0.$$

Thus applying De-Morgans Law we can say, $$P(\bigcap_{\epsilon>0 \ rationals \ } \bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} \{ |Y_n-Y| < \epsilon\}) = 1.$$

This implies $$P( \bigcup_{\epsilon > 0 \ rationals}\bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} \{ |Y_n-Y| < \epsilon\}) = 1.$$

We can also note $$\bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} \{ |Y_n-Y| < \epsilon\} \subset \bigcap_{m=1}^{\infty} \bigcup_{n = m}^{\infty} \{|Y_n-Y|> \epsilon \}$$.

Is this enough to show it converges almost surely? I remember being told you also have to show the two sets on both the LHS and RHS have the same elements? If so how would you go about doing this?

• You've basically gotten it. If it helps there's a nice application of Borel Cantelli which says that if there exists a monotone decreasing sequence $\epsilon_n$ such that $\sum_nP(|Y_n-Y|>\epsilon_n)<\infty$ then you have almost sure convergence. You can easily extract such a sequence from your assumptions – Alex R. May 6 '16 at 17:19

since $\sum_{j=1}^{\infty} P(\{ Y_j > \epsilon \}) < \infty$. In the second step we've used the fact that $\cup_{j=i}^{\infty} \{ Y_j > \epsilon \}$ is monotone decreasing in $i$ and so we can write the intersection as a limit which can be taken outside the probability by the monotone convergence theorem. The second to last is of course Boole's inequality.