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.

  • $\begingroup$ 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? $\endgroup$ – Von Kar Apr 13 '16 at 12:12
  • $\begingroup$ Where's the typo? It seems ok to me. $\endgroup$ – carmichael561 Apr 13 '16 at 15:47
  • $\begingroup$ 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. $\endgroup$ – Von Kar Apr 13 '16 at 15:56

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.