Example 7.4.4(e): Lebesgue Integral for Simple Functions
First assume that there are two representations for s
sA(x) = aj XAj(x) with Aj disjoint and aj not zerosuch that sA(x) = sB(x). Let's assume that the integral of s using the first representation exists, i.e. the measure of all sets Aj is finite.
sB(x) = bk XBk(x) with Bk disjoint and bk not zero
If x Bk then x must be contained in one of the Aj's because otherwise sB(x) # 0 and sA(x) = 0. Therefore Bk Aj so that m(Bk) is finite for all k so that sB is integrable.
By the same reasoning we have
Bk = j ( Aj Bk ) and Aj = k ( Bk Aj )If Aj Bk is not empty, then for x Aj Bk we have
aj = sA(x) = sB(x) = bkso that aj = bk in that case. Putting everything together gives:
so that the integrals agree regardless of the representation.
It remains to show that the integral of a simple function agrees with the integral over the canonical representation of that function, i.e. when the sets are disjoint and the coefficients are not zero.