# $\int_{\bigcup_{n=1}^{\infty}E_n}f=\sum_{n=1}^{\infty}\int_{E_n}f$ given $f$ positive and measurable

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.

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.

• 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? – glpsx Apr 13 '16 at 12:12
• Where's the typo? It seems ok to me. – carmichael561 Apr 13 '16 at 15:47
• 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. – glpsx Apr 13 '16 at 15:56

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.