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Why $EN < \infty $ implies $ N < \infty $? (It's part of Borel-Cantelli lemma's proof).

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What is $E$ and $N$? (In particular, could any of them fail to be finite and if so how?) –  Henning Makholm Sep 30 '11 at 19:42

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up vote 6 down vote accepted

If $\mathbb{E}[N] < \infty$, then necessarily $\mathbb{E}[N^+] < \infty$ and $\mathbb{E}[N^-] < \infty$, where $N^+(\omega): = \max\{N(\omega),0\}$ and $N^-(\omega): = \max\{-N(\omega),0\}$, the positive part and negative part of $N$ respectively. This is because of how expectation is defined, as the sum of the expectations of the positive part and negative part. Now, If $N$ (and thus $N^+$) took the value positive infinity on a set of positive measure $A$, then $\mathbb{E}N^+ \geq \mathbb{E}[N^+ \chi_A] = \infty$.

So since the expectation of $N$ is finite, the set where $N$ is infinite must have zero measure, or in other words, $N < \infty$ a.s. The same argument applies to $N^-$, implying that the $N > -\infty$ a.s. as well.

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Thanks. BTW, what does "a.s." mean? –  ablmf Sep 30 '11 at 20:22
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a.s. is almost surely, see en.wikipedia.org/wiki/Almost_surely for full description. –  karakfa Sep 30 '11 at 20:28
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It is not true in general that $N \geq \chi_A$. But, $\mathbb{E}N < \infty \Rightarrow \mathbb{E}N^+ < \infty$, and $N^+ \geq \chi_A$. Maybe you should observe that. –  André Caldas Oct 14 '11 at 13:53
    
Thank you Andre, you are right. I have edited my answer to include this. –  Jeremy Voltz Oct 17 '11 at 18:45

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