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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!

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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 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.

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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

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