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I'm learning about measure theory (specifically Lebesgue intregation) and need help with the following problem:

Let $f:\mathbb{R}\rightarrow[0,+\infty)$ be measurable and let $\{E_n\}$ be a collection of pairwise disjoint measurable sets. Prove that $\int_{\bigcup_{n=1}^{\infty}E_n}f=\sum_{n=1}^{\infty}\int_{E_n}f.$

For convenience I set $E=\bigcup_{n=1}^{\infty}E_n$.

This problem looks like an application of the monotone convergence theorem but I'm having a hard time applying it. I need to find a sequence of functions that is positive an nondecreasing but I don't know how to define it.

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Let $E=\bigcup_{n=1}^{\infty}E_n$, then $f\chi_E=\sum_{n=1}^{\infty}f\chi_{E_n}$, hence $$ \int_Ef=\int_{\mathbb{R}}f\chi_E=\int_{\mathbb{R}}\sum_{n=1}^{\infty}f\chi_{E_n}=\sum_{n=1}^{\infty}\int_{\mathbb{R}}f\chi_{E_n}=\sum_{n=1}^{\infty}\int_{E_n}f$$ The monotone convergence theorem is what allows us to interchange the sum and integral, with $g_m=\sum_{n=1}^mf\chi_{E_n}$ being the non-decreasing sequence.

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  • $\begingroup$ Thank you for your clear explanation. I think there is a typo in your last sentence with the indexes in the sum. Could you verify it so I can accept your answer? $\endgroup$ – glpsx Apr 13 '16 at 12:12
  • $\begingroup$ Where's the typo? It seems ok to me. $\endgroup$ – carmichael561 Apr 13 '16 at 15:47
  • $\begingroup$ You are correct, sorry about that. I got confused by the fact that we are summing from $n = 1$ to $m$ for $g_m$. Thank you for your clear explanation (and time), much appreciated. $\endgroup$ – glpsx Apr 13 '16 at 15:56
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Set $f_N=\sum_{n=1}^{N} f\chi_{E_n}$.

As $f\chi_{E_n}\geq 0$ for each $n$, $f_1\leq f_2 \leq f_3 \leq f_4....$.

Now observe that $f_n \rightarrow f \chi_{E}$ as $n\rightarrow \infty$, where $E=\bigcup_{n=1}^{\infty}E_n$. Thus by MCT we obtain the following equality:

$\lim_{n} \int f_n d\mu= \int f\chi_{E} d\mu$.

Which is the conclusion you seek.

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