Let $ A_n =\lbrace x \in \Omega : n+1 \geq |f(x)| > n \rbrace$ for $f$ an integrable function over an open set $\Omega$ included in $ \mathbb{R}^n $.

How can you show that $ \sum n|A_n| $ converges ?

You can use the fact that for all $\epsilon > 0 $, there exist a set $E$ of measure $ \leq \epsilon$ and such that $f$ is bounded over the complementary of $E$. You prove that with Markov's inequality.

  • $\begingroup$ What is $|A_n|$? Do you mean $\mu(A_n)$, the measure of the set? $\endgroup$ – B. Mehta May 17 '18 at 16:54
  • $\begingroup$ Yes, I meant that. $\endgroup$ – MrMaths May 17 '18 at 16:55
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    $\begingroup$ The answer here should help. $\endgroup$ – B. Mehta May 17 '18 at 17:17
  • $\begingroup$ Yes, it did @B.Mehta $\endgroup$ – MrMaths May 17 '18 at 17:18
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    $\begingroup$ You are using $n$ for too many things. $\endgroup$ – zhw. May 17 '18 at 17:46

I don't think you stated the question correctly. If we consider $f(x)=\frac{1}{\sqrt{x}}$ over $(0,1)$. Then $A_n=(0,\frac{1}{n^2})$ and $\sum_{n=1}^{\infty}n |A_n|$ is infinity.

  • $\begingroup$ You are right. I corrected the mistake. $\endgroup$ – MrMaths May 17 '18 at 17:08
  • $\begingroup$ you can consider $\cup_{n=0}^{\infty} A_n\subset \Omega$, and consider the $L^1$ norm on these two measurable sets. $\endgroup$ – Ben May 17 '18 at 17:13

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