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?

  • $\begingroup$ 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 $\endgroup$
    – Alex R.
    May 6, 2016 at 17:19

1 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.

  • $\begingroup$ Although true, you haven't shown that complete convergence is stronger than almost sure convergence. You only showed, as was sufficient to answer the posted question, that complete convergence implies almost sure convergence. $\endgroup$ May 6, 2016 at 16:43
  • $\begingroup$ @MarkL.Stone Isn't that what it means for one condition to be stronger than another? $\endgroup$
    – dsaxton
    May 6, 2016 at 18:11
  • $\begingroup$ No, it's not. You showed that complete convergence implies almost sure convergence. You did not show that almost sure convergence does not imply complete convergence. I.e., you did not show that complete convergence and almost sure convergence are not equivalent. $\endgroup$ May 6, 2016 at 18:20

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