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}^d $.
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.
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.