Let (X,Θ,µ) be a measure space, and suppose X = n∈Z+ En, where {En} is a pairwise disjoint collection of measurable subsets of X. Use the monotone convergence theorem and linearity of the integral to prove that, if f is a non-negative measurable real-valued function on X, then ∞ f dµ = X n=1En f dµ.

Answer 1

Consider the sequence of functions {fn} defined by fn(x) = f(x) if x ∈ E1 ∪ E2 ∪ ... ∪ En and fn(x) = 0 otherwise.

Note that each fn is non-negative and measurable since f is non-negative and measurable, and each En is measurable.

Moreover, fn(x) → f(x) as n → ∞ for all x in X, since for any x ∈ E1∪E2∪...∪En for some large enough n, we have fn(x) = f(x) and for all m > n, we have fm(x) = 0.

Therefore, by the monotone convergence theorem, we have:

∞ ∫ f dµ = lim ∫ fn dµ
n→∞ X n=1 En ∫ f dµ = X n=1 En ∫ fn dµ

But each En ∫ fn dµ is just the integral of f over the set En, since fn is zero outside of E1 ∪ E2 ∪ ... ∪ En. Therefore, we have:

∞ ∫ f dµ = X n=1 En ∫ f dµ

which is the desired result.