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Let $(X,\Sigma,\mu)$ a measure space and $f\in L_p$, where $p\in [1,+\infty)$. Let $(A_n)$ be a sequence in $\Sigma$ such that $\mu(A_n)\to 0$. Then I want to prove that $\int_{A_n}fd\mu\to 0$.

I thought I solved this but this is something strange.

First we suppose that $f:=\chi_{E}$ for some $E\in\Sigma$. Then $\int _{A_n}fd\mu=\mu(A_n\cap E)$ who clearly converges to zero.

It is obvious now that the statement holds for non-negative and measurable simple-functions.

Now if $f$ is non-negative and measurable, there exists $(s_m)$ a sequence of simple-measurable functions such that $s_m\to f$. By Monotone Convergence Theorem $$\lim_{n\to\infty}\int_{A_n}f=\lim_{m\to\infty}\lim_{n\to\infty}\int_{A_n}s_m=0.$$

Now, if $f$ is measurable, then $\int_{A_n}f^+\to 0$ and $\int_{A_n}f^-\to 0$, and the statement holds.

Of course I think there is something wrong, as I didn't use that $f\in L_p$. Where is it?

Thank you.

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  • $\begingroup$ In general, you can't interchange limits. $\endgroup$ – kobe Apr 25 '15 at 23:47
  • $\begingroup$ @kobe even if a sequence $a_{n,m}\to 0$? $\endgroup$ – Sandor Apr 25 '15 at 23:52
  • $\begingroup$ That would be different but this has not been established in your argument. $\endgroup$ – kobe Apr 25 '15 at 23:55
  • $\begingroup$ @kobe I see. Most likely it can't be established, can it? Otherwise we would not use $f\in L_p$ and there would be something strange as I thought. $\endgroup$ – Sandor Apr 26 '15 at 0:08
  • $\begingroup$ Right, it's usually a red flag when you don't use all the conditions. :) $\endgroup$ – kobe Apr 26 '15 at 0:14
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Consider using Holder's inequality with conjugate exponents $p$ and $p/(p-1)$:

$$\left|\int_{A_n} f\, d\mu\right| \le \int_X \chi_{A_n}|f|\, d\mu \le \|\chi_{A_n}\|_{p/(p-1)} \|f\|_p = \mu(A_n)^{(p-1)/p} \|f\|_p.$$

Since $\mu(A_n) \to 0$, the result follows.

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  • $\begingroup$ Why is the inequality $\left|\int_{A_n} f\, d\mu\right| \le \int_X \chi_{A_n}|f|\|$ valid if we don't know if $f \in L_1$? $\endgroup$ – Twink Apr 13 at 22:03

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