Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I have a question concerning integration theory I can't figure out, maybe someone can help:

Fix $\varepsilon>0$ and consider $\delta \colon [0,1] \to (0,\infty)$ measurable. Is it then true that $$\inf_{N\subset [0,1], \lambda(N)>\varepsilon} \int_N \delta(t) dt > 0$$ where $\lambda(N)$ is the Lebesgue measure of $N$?

Thank you!

share|improve this question
1  
Hello Stano, and welcome to MO. To me your question does not seem to be a research-level question (see faq) so in the future I would recommend the site math.stackexchange.com for questions like these. –  Rami Luisto Jan 24 '13 at 12:55

1 Answer 1

up vote 1 down vote accepted

If it wasn't the case, we would be able to find measurable sets $N_k$ such that $\lambda(N_k)>\varepsilon$ and the integral on $N_k$ of $f$ is smaller than $k^{-1}$. So the sequence $\{f\chi_{N_k}\}$ converges in $L^1$ to $0$ (provided $f$ is integral, which can be assumed), and we extract a subsequence $\{f\chi_{S_k}\}$ which converges almost everywhere to $0$. So $\displaystyle\mu(\limsup_{k\to +\infty}S_k)=0$ and we can find an integer $j_0$ such that $\displaystyle \mu(S_{j_0})\leqslant \mu\left(\bigcup_{n\geqslant j_0}S_n\right)<\varepsilon/2$, a contradiction.

share|improve this answer
    
Thanks a lot! it took me a while to understand it fully, but now I get it. Great –  Stano Jan 24 '13 at 20:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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