Let $(X,F,\mu)$ be a measure space and $P\subseteq Q\in F$. Let $f:X\rightarrow\mathbb{R}$ be a measurable function such that $f(x)=0$ for all $x\in Q-P$ (the minus denotes set difference). Is there any easy way to check that $\int_P fd\mu=\int_Q fd\mu$? I think it would be routine but tedious to check beginning from the definition (i.e. integration of simple functions.) Does it follow easily from some property/theorem of Lebesgue integral?

  • $\begingroup$ $\int_Q fd\mu = \int_P fd\mu+ \int_{ (Q-P)} fd\mu $ $\endgroup$ – DBFdalwayse Sep 28 '13 at 2:12

I think this should work: $\int_Q fd\mu = \int_P fd\mu+ \int_{ (Q-P)} fd\mu $, and you know f is $0$ on $Q-P $

  • $\begingroup$ How would you prove that $\int_Q fd\mu = \int_P fd\mu+ \int_{ (Q-P)} fd\mu$ from first principles? $\endgroup$ – Keshav Srinivasan Sep 28 '13 at 2:19
  • $\begingroup$ Well, if two sets $A,B$ are disjoint, then $m(A\cup B)$=$m(A)+m(B)$ $\endgroup$ – DBFdalwayse Sep 28 '13 at 3:48

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.