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Here is the following problem:

Suppose that $f$ is Lebesgue integrable over E and that $\{E_{n}\}_{n=1}^{\infty}$ is a increasing sequence of lebesgue measurable sets, that is, $E_1$ $\subset$ $E_2$ $\subset \cdots$ and $\cup_{n=1}^{\infty} E_{n}$ := E. Prove that

$$\int_E f\,d\lambda = \lim_{n \to \infty} \int_{E_n} f\,d\lambda$$

Any suggestions/ hints on how to start this problem would be appreciated.

Community
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FoxViking
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2 Answers2

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Hint: Let $f_n := f\chi_{E_n}$, where $\chi$ is the characteristic function. Then $f \chi_E = \lim_{n\to\infty} f_n$. Hence you would like to move the limit inside the integral, which theorem would be useful for that?

I was suspended for talking
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If $\nu(A):=\int_A f_+d\lambda$ and $\mu(A):=\int_A f_-d\lambda$ then $\nu$ and $\mu$ are both finite measures so that: $$\lim_{n\to\infty}\nu(E_n)=\nu(E)<\infty\text{ and }\lim_{n\to\infty}\mu(E_n)=\mu(E)<\infty$$ So: $$\lim_{n\to\infty}\int_{E_n} fd\lambda=\lim_{n\to\infty}[\nu(E_n)-\mu(E_n)]=\nu(E)-\mu(E)=\int_Efd\lambda$$

drhab
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  • From a logical point of view, there is a slight problem with this approach, which is that, to prove that nu and mu are measures, one needs to show that the desired result holds. – Did Mar 19 '18 at 07:30
  • @Did Yes, you are right. This cannot be marked as a complete answer. But it makes things less complex: it is enough to show that the desired result holds for non-negative $f$. – drhab Mar 19 '18 at 10:16