Here is the problem: If $f$ is a nonnegative Lebesgue measurable function and $\{E_n\}_{n=1}^{\infty}$ is a sequence of Lebesgue measurable sets with $E_1\subset E_2\subset ...$, then

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

I know that $\int_{\bigcup \limits_{n=1}^{\infty} E_{n}} f \,d\lambda = \int \limits_{\Bbb R} f \chi_{\bigcup \limits_{n=1}^{\infty} E_{n}} \,d\lambda$.

I'm also thinking that I should apply the Monotone Convergence Theorem.

However, I don't know where to go from here. Any help would be appreciated.

  • 2
    $\begingroup$ Possible duplicate of Not sure how to start proving that $\int_E f\,d\lambda$ = $\lim_{n \to \infty} \int_{E_n} f\,d\lambda$ $\endgroup$ – amWhy Feb 25 '18 at 0:09
  • 1
    $\begingroup$ DO NOT ask the same question TWICE. $\endgroup$ – amWhy Feb 25 '18 at 0:11
  • $\begingroup$ How could you not know it was already asked, when you asked the question I link, and ask it again here?: NOT okay. $\endgroup$ – amWhy Feb 25 '18 at 0:45
  • $\begingroup$ "Here is the following problem: Suppose that f is Lebesgue integrable over E and that $\{E_{n}\}_{n=1}^{\infty}$ is a sequence of lebesgue measurable sets with $E_1$ $\subset$ $E_2$ $\subset$.... and $\cup_{n=1}^{\infty} (E_{n})$ = E. Prove that $\int_E f\,d\lambda$ = $\lim_{n \to \infty} \int_{E_n} f\,d\lambda$. Any suggestions/ hints on how to start this problem would be appreciated." Asked already by you here $\endgroup$ – amWhy Feb 25 '18 at 0:48
  • $\begingroup$ Hold up....these are two different questions. I simply thought you were referring to a different question (that was similar to this question) asked by someone else. $\endgroup$ – FoxViking Feb 25 '18 at 0:49

Write $\displaystyle\int_{\bigcup_{n=1}^{\infty}E_{n}}fd\lambda=\int_{X}\chi_{\bigcup_{n=1}^{\infty}E_{n}}fd\lambda$ and $\displaystyle\int_{E_{n}}fd\lambda=\int_{X}\chi_{E_{n}}fd\lambda$ and now observe that $f\chi_{E_{n}}\uparrow f\chi_{\bigcup_{n=1}^{\infty}E_{n}}$ so Monotone Convergence Theorem applies here.

  • $\begingroup$ $f\chi_{E_{n}}\uparrow f\chi_{\bigcup_{n=1}^{\infty}E_{n}}$ follows since f is a nonnegative Lebesgue measurable function and $\{E_n\}_{n=1}^{\infty}$ is a monotone nondecreasing sequence of Lebesgue measurable sets? $\endgroup$ – FoxViking Feb 24 '18 at 23:52
  • $\begingroup$ Yes, you are right. $\endgroup$ – user284331 Feb 24 '18 at 23:54
  • $\begingroup$ Alright. Thanks! $\endgroup$ – FoxViking Feb 24 '18 at 23:56

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.