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Let $\left(X,M,\mu\right)$ be a measurable space and $f_{k}:X\rightarrow \left[0,\infty\right]$ with $k \in \mathbb{N}$ a decreasing succession of measurable functions with $\int_{X}f_{k}d\mu<\infty$ for all $k \in \mathbb{N}$. Show that $$\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu=\int_{X}\lim_{k\rightarrow\infty}f_{k}d\mu$$

Remark: I have tried the following:

We know that if $f_{k}\geq f_{k+1}$ then $\int_{X}f_{k}d\mu \geq \int_{X}f_{k+1}d\mu$. Therefore: $$\liminf_{k\rightarrow\infty}f_{k}=\lim_{k\rightarrow\infty}f_{k}\; \;\mbox{ y }\; \; \liminf_{k\rightarrow\infty}\int_{X}f_{k}d\mu=\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu$$ Hence, by Fatou's Lemma we have $$\int_{X}\lim_{k\rightarrow\infty}f_{k}d\mu\leq\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu$$ The problem is to prove the other inequality.

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  • $\begingroup$ This is immediate from the Dominated Convergence Theorem. $\endgroup$ Commented Oct 6, 2015 at 1:26
  • $\begingroup$ @DavidC.Ullrich Thanks, you're absolutely right. $\endgroup$ Commented Nov 3, 2015 at 3:12

1 Answer 1

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The other direction is the "easy" direction: it follows directly from definitions.

Write out the definition of the integral as the supremum of the integrals over simple functions below $f$. Now what can you say about the integral of each $f_k$ compared to the integral of $f$?

This is a common feature in Lebesgue integration arguments. Equalities are proved using two inequalities, and one inequality is usually free from the definitions.

p.s. Don't forget you need to address measurability of $\lim_{k\rightarrow\infty} f_k$ before you can do anything.

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