# Extending Lebesgue integral with zero value

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?

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