0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.