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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?

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

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  • $\begingroup$ 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

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