# measurable and finite almost everywhere

Let $f:[a,b] \rightarrow [-\infty, \infty]$ be measurable and finite almost everywhere. Show that given $\epsilon >0$ there exists $0<M<\infty$ so that $m(\{|f|>M\})<\epsilon$.

My works.

I was trying to prove my contradiction, but I couldn't. I know that $m(\{f=\infty\})=o$, since its finite measure.

Are there anyway to solve this problem? Thanks.

Also, What would happen if we replace $[a,b]$ by a more general set $E\subset$ and still get the same conclusion?? Thanks

Hint: Let $E_n=\{x \in [a,b]: |f(x)| \ge n\}$. Then $E_n$ is measurable. Moreover $E_{n+1} \subset E_n$ for all $n \in \mathbb{N}$. Then $$m(\cap_{n=1}^{\infty}E_n)=\lim_{n \to \infty}E_n$$
Show that $$\cap_{n=1}^{\infty}E_n=\{x \in [a,b]: |f(x)|=\infty\}$$ and conclude from here.
• How do we know measure of $E_1$ is finite? Does measurable imply finite? Like if we did this on $\mathbb{R}$ instead of $[a,b]$. – user389056 Dec 16 '19 at 17:10