# How does complete convergence imply almost sure covergence?

Let $Y_1$, $Y_2$,... be a sequence of random variables on a probability space ($\Omega$ , F , $\mu$). The definition of complete convergence is for $\epsilon$>0, $$\sum_{n=1}^\infty P\{\lvert Y_n-Y \lvert > \epsilon \} < \infty.$$

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 $$\bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} \{ |Y_n-Y| < \epsilon\} \le \bigcap_{m=1}^{\infty} \bigcup_{n = m}^{\infty} \{|Y_n-Y|< \epsilon \}.$$

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?

I think your statement $P( \bigcup_{\epsilon>0 \ rationals} \bigcap_{m=1}^{\infty} \bigcup_{n > m}^{\infty} \ \{ |Y_n - Y| < \epsilon \}) = 1$ for almost sure convergence is mistaken, and should be $P( \bigcap_{\epsilon>0 \ rationals} \bigcup_{m=1}^{\infty} \bigcap_{n > m}^{\infty} \ \{ |Y_n - Y| < \epsilon \}) = 1$, which takes care of your problem (your applications of Borel-Cantelli and De Morgan are enough to do the job).