Show that $Y_n$ converges to Y almost surely

Question

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?

Answer 1

You can prove this using Boole's inequality alone.

\begin{align} P( \{ |Y_n - Y| > \epsilon \text{ i.o.} \}) &= P(\cap_{i=1}^{\infty} \cup_{j=i}^{\infty} \{ Y_j > \epsilon \}) \\ &= P(\lim_{i \to \infty} \cup_{j=i}^{\infty} \{ Y_j > \epsilon \}) \\ &= \lim_{i \to \infty} P(\cup_{j=i}^{\infty} \{ Y_j > \epsilon \}) \\ &\leq \lim_{i \to \infty} \sum_{j=i}^{\infty} P(\{ Y_j > \epsilon \}) \\ &= 0 \end{align}

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.

This kind of convergence is sometimes called "complete convergence" and as you can see it's stronger than almost sure convergence.