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The problem I am referring to is Problem 40 of the following document:

https://huynhcam.files.wordpress.com/2013/07/anhquangle-measure-and-integration-full-www-mathvn-com.pdf

Let $(f_n)$ be a sequence of extended real-valued measurable functions on $X$ and let $f$ be an extended real-valued function which is finite a.e. on $X$. Suppose $\lim_{n\to\infty}f_n=f$ a.e. on $X$. Let $\alpha\in[0,\mu(X))$ be arbitrarily chosen. Show that for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that $\mu\{X:|f_n-f|<\varepsilon\}\geqslant \alpha$ for $n\geqslant N$.

I can't understand where they pulled the first equation from. At first it seemed to be an application of the Borel-Cantelli Lemma (mentioned a couple of pages back) but this would require showing that the series converges, but I don't see how to do this and it seems like a strictly stronger statement than what we're trying to prove (that the terms of that series converge to zero). I can't understand how this is 'obvious' if it is not some application of a lemma. The rest of the solution clearly follows, at least after the application of Fatou's Lemma in the next line. So it seems the crux of the proof has simply been stated. Any explanation or even just a hint would be appreciated.

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  • $\begingroup$ Did you reference the footnote? $\endgroup$ – Math1000 Dec 13 '18 at 20:14
  • $\begingroup$ Perhaps I should've mentioned it, but then again it is just explaining that the second equality follows from the first by Fatou's Lemma, which I understand. But it doesn't help me understand where the first equality came from. $\endgroup$ – AlephNull Dec 13 '18 at 20:38
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    $\begingroup$ Try writing the definition of $\limsup$ and comparing it to the definition of convergence a.e. $\endgroup$ – Math1000 Dec 13 '18 at 20:41
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    $\begingroup$ I am not really sure what I was thinking since I only just realised it was a lim sup of sets. Now it is obvious as it should have been. $\endgroup$ – AlephNull Dec 13 '18 at 21:38

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