Example 7.4.4(e): Lebesgue Integral for Simple Functions

We have seen before that the representation of a simple function is not unique. Show that the Lebesgue integral of a simple function is independent of its representation.
Context

We have to show that if s is a simple function with two different representations, then the integrals using either representation agree.

First assume that there are two representations for s

s_A(x) = a_j X_{A_j}(x) with A_j disjoint and a_j not zero
s_B(x) = b_k X_{B_k}(x) with B_k disjoint and b_k not zero

such that s_A(x) = s_B(x). Let's assume that the integral of s using the first representation exists, i.e. the measure of all sets A_j is finite.

If x B_k then x must be contained in one of the A_j's because otherwise s_B(x) # 0 and s_A(x) = 0. Therefore B_k A_j so that m(B_k) is finite for all k so that s_B is integrable.

By the same reasoning we have

B_k = _j ( A_j B_k ) and A_j = _k ( B_k A_j )

If A_j B_k is not empty, then for x A_j B_k we have

a_j = s_A(x) = s_B(x) = b_k

so that a_j = b_k 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.

Interactive Real Analysis, ver. 1.9.5
(c) 1994-2007, Bert G. Wachsmuth
Page last modified: Mar 28, 2007