# Lebesgue integral over unions

I want to prove the following:

Let $f$ be a nonnegative $\mathcal{M}$-measurable function, and let $\{E_n\}_{n=1}^{\infty}$ be a sequence of Lebesgue measurable sets, where $E_1\subset E_2\subset ...$

Then $$\int_{\bigcup_{n=1}^{\infty}E_n}f{\rm d}\lambda = \lim_{n\to\infty}\int_{E_n}f{\rm d}\lambda$$

Can anyone help? I am very new to measure theory, and don't manage to prove this on my own.

• hint: monotone convergence theorem – zhw. Nov 4 '16 at 17:02

$\int_{\bigcup \limits_{n=1}^{\infty} E_{n}} f \,d\lambda = \int \limits_{X} f \chi_{\bigcup \limits_{n=1}^{\infty} E_{n}} \,d\lambda$ by definition of integrating over a subset of the domain $X$.
Also, the characteristic function $\chi_{\bigcup \limits_{n=1}^{\infty} E_{n}}$ can be written as $\lim \limits_{m \to \infty}\chi_{\bigcup \limits_{n=1}^{m} E_{n}}$, right?
Also, $\bigcup \limits_{n=1}^{m} E_{n} = E_{m}$ (why?). So, you want to show:
$\int \limits_{X} f ( \lim \limits_{m \to \infty} \chi_{E_{m}}) \,d\lambda = \lim \limits_{m \to \infty} \int \limits_{X} f \chi_{E_{m}} \,d\lambda$. Now look at user zhw.'s hint.