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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?

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    $\begingroup$ You are not supposed to append an answer to your question as if it was a part of the question. $\endgroup$ – Did Feb 27 '16 at 9:44
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Since $\sum_{n=1}\mathbb P(|Y_n-Y|>\varepsilon)<\infty$, the Borel-Cantelli lemma implies that $$\mathbb P\left(\limsup_{n\to\infty} |Y_n-Y|>\varepsilon\right) = 0. $$ From this we conclude that $Y_n\to Y$ a.s.

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