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

Question

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.

Answer 1

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.

Answer 2

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.