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Let $(X, M, \mu)$ be a measurable space. I'm trying to prove the following statement:

If $f \in L^p$, $1<p<\infty$ and $\{A_n\}$ is a sequence of measurable sets sucht that $\mu(A_n)\to 0$ then $\int_{A_n} f d\mu \to 0$.

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Maybe that (I assumed $f$ positive, but by separating $f = f^{+} - f^{-}$ it should be the same proof) :

Assume the opposite, then it exists $\varepsilon > 0$ such that $\forall N \geq 0: \exists n \geq N: \int_{A_n}f d \mu > \varepsilon$.

Now using Hölder inequality ($q$ is such that $\frac{1}{p} + \frac{1}{q} = 1)$, $\int_{A_n} f \,\mathrm{d}\mu = \int_{X} f \cdot 1_{A_n} \,\mathrm{d}\mu \leq \lVert f\rVert_p \cdot \left(\mu(A_n)\right)^{q}$ but this gives us $\mu(A_n) \geq \left(\frac{\varepsilon}{\lVert f\rVert_p}\right)^{\frac{1}{q}} > 0$ which contradicts the fact that $\mu(A_n) \rightarrow 0$.

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  • $\begingroup$ very nice proof the only problem is that the question did not state that the function is real valued $\endgroup$ Apr 16 '14 at 6:17
  • $\begingroup$ you're right, but it should be easily adaptable for $f$ with values in $\mathbb{C}$ or any finite dimensionnal $\mathbb{R}$-vectorial space when the integral is just generalized as the integral of each coordinate functions. If you talk about a function with values in a Hilbert space or a Banach space, the prof doesn't work anymore indeed but this is another story, I assumed that he uses here the usual Lebesgue integral...maybe coming back to the definition, showing the result for simples functions as usual and so on work for both cases but I find this way more elegant and less painful ^^ $\endgroup$
    – yago
    Apr 16 '14 at 11:07

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