3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ hint: monotone convergence theorem $\endgroup$ – zhw. Nov 4 '16 at 17:02
0
$\begingroup$

Here are some additional hints:

$\int_{\bigcup \limits_{n=1}^{\infty} E_{n}} f \,d\lambda = \int \limits_{X} f \chi_{\bigcup \limits_{n=1}^{\infty} E_{n}} \,d\lambda$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.