# If $\mu(A_n)\to 0$ then $\int_{A_n} f d\mu \to 0$.

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

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

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 $$\mu(A_n) \rightarrow 0$$.
• 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 ^^