Does the following fact have a name?

**Theorem 1** * Suppose are a countable collection of 0/1 random variables over a probability space such that for every integer and bits the event*

*
*
is measurable.

Suppose that

Then

* *

*Proof:* Intuitively, we want to say that by linearity of expectation we have and so by Markov’s inequality

Just to make sure that this can be made rigorous, let’s belabor the proof step by step, doing everything completely from first principles.

Let .

First of all, the event

is measurable, because it is the countable intersection over all of the events , which are measurable because each of them is the countable union over all of the events . So suppose towards a contradiction that its probability is not zero, then there is such that

In particular, for every ,

But, by linearity of expectation, Markov’s inequality, and our assumption, we have that for every and every

Now,

which is a contradiction if we choose .

Maybe we should also justify . Define the disjoint events as

Then

and

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## 12 comments

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December 23, 2009 at 5:19 pm

BatzI needed something very similar in a recent paper, and we showed it followed directly from the central limit theorem (we used a version for independent non-identically distributed random variables, see pg 176 in Grimmett “Probability and random processes”).

December 23, 2009 at 5:22 pm

commenterborel-cantelli

December 23, 2009 at 5:30 pm

harrisonThis is exactly the sort of thing MathOverflow is excellent for.

December 23, 2009 at 5:31 pm

harrisonAhh, wordpress ate my fake HTML tag. So, append [/shameless plug] to the above.

December 23, 2009 at 5:33 pm

anonconfirm, Borel-Cantelli Lemma

December 23, 2009 at 5:41 pm

lucaThanks, commenter.

Batz, note that the statement is true even if you don’t have independence, which makes it much more versatile.

Harrison: yes, after posting it here it occurred to me that mathoverflow would have been a better place. But I still got an answer in 27 minutes.

December 23, 2009 at 5:52 pm

pierreIndeed, Borel-Cantelli.

Moreover, your proof seems needlessly complicated. All you need to show is

that

E[ \sum_i X_i] < infinity,

which would immediately rule out the possibility that \sum_i X_i = infinity

with positive probability. But this follows right away from interchanging sum

and expectation; the interchange can be easily justified since each X_i is nonnegative.

December 24, 2009 at 3:56 pm

Jonathan Vos PostThis comment no verb.

January 4, 2010 at 7:47 pm

anonLuca, is the big news public yet?

January 4, 2010 at 9:43 pm

reply to anonWhat is the big news? Who proved what?

January 5, 2010 at 5:47 pm

anonLuca proved himself right, I hear.

January 11, 2010 at 4:00 pm

reply to anonLuca to Stanford ?! That’s certainly unexpected..