Show that $\sum n|A_n|$ converges

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.

• What is $|A_n|$? Do you mean $\mu(A_n)$, the measure of the set? – B. Mehta May 17 '18 at 16:54
• Yes, I meant that. – MrMaths May 17 '18 at 16:55
• The answer here should help. – B. Mehta May 17 '18 at 17:17
• Yes, it did @B.Mehta – MrMaths May 17 '18 at 17:18
• You are using $n$ for too many things. – 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.
• you can consider $\cup_{n=0}^{\infty} A_n\subset \Omega$, and consider the $L^1$ norm on these two measurable sets. – Ben May 17 '18 at 17:13