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Let $(X,\mathcal F, \mu)$ be measurable space with $\mu(X)<\infty$. Prove that if function $f$ is measurable and finite on $X$ then $$\lim_{n\to \infty}\mu \{x: |f(x)|\geq n\}=0.$$

I have been asked this question before. However, I missed one condition that $f$ is finite. However, I got a hint without using this hypothesis and I believe that the hint is insufficient. Can anyone give me more detail proof? Thanks

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    $\begingroup$ By finite do you mean $f(x) < \infty$ for all $x \in X$? Seems like some people have interpreted finiteness as boundedness in their answers. $\endgroup$ – desos Aug 20 '15 at 14:04
  • $\begingroup$ Yes,you're right $\endgroup$ – Hoan Do Aug 20 '15 at 14:11
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    $\begingroup$ and 3 deleted answers later, we finally have a correct proof :) $\endgroup$ – air Aug 20 '15 at 14:16
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Perhaps, this argument is easier. Just notice the identity $$(|f|=+\infty)=\bigcap_{n\geq 1} (|f|\geq n).$$ Hence $$0=\mu(|f|=+\infty)=\mu(\bigcap_{n\geq 1}(|f|\geq n))=\lim_n\mu(|f|\geq n),$$ where the second equality is continuity from above because $\mu$ is finite.

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