# if $\mu(X)$ is finite and $f$ is finite on X a.e then $\lim_{n\to \infty}\mu \{x: |f(x)|\geq n\}=0$

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

• 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. – desos Aug 20 '15 at 14:04
• Yes,you're right – Hoan Do Aug 20 '15 at 14:11
• and 3 deleted answers later, we finally have a correct proof :) – air Aug 20 '15 at 14:16

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.