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.
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.
To prove that ∫∞ f dµ = ∑n=1∞ ∫En f dµ, we will use the monotone convergence theorem and linearity of the integral.
Let A1 = E1, A2 = E1 ∪ E2, A3 = E1 ∪ E2 ∪ E3, and so on.
Note that {An} is an increasing sequence of sets, meaning that An ⊆ An+1 for all n. Also, X = ∪n=1∞ An, as given in the problem.
Now, consider the sequence of functions {fn} defined as follows:
fn(x) = f(x) if x ∈ En, and fn(x) = 0 if x ∉ En.
It is clear that fn is a non-negative measurable real-valued function on X, as fn(x) = 0 for x ∉ En and fn(x) = f(x) for x ∈ En.
Furthermore, for any fixed x ∈ X, fn(x) ≤ fn+1(x) for all n, since En ⊆ En+1 for every n. Therefore, {fn} is an increasing sequence of functions.
By the monotone convergence theorem, we have:
∫∞ f dµ = ∫X lim n→∞ fn dµ.
Now, let's use linearity of the integral:
∫X lim n→∞ fn dµ = lim n→∞ ∫X fn dµ.
Since fn(x) = 0 for x ∉ En, we can rewrite the integral as follows:
lim n→∞ ∫X fn dµ = lim n→∞ ∫En fn dµ.
But the integral over En is simply the integral of f restricted to En, since fn(x) = f(x) for x ∈ En. Therefore, we have:
lim n→∞ ∫En fn dµ = lim n→∞ ∫En f dµ.
Now, notice that ∫∞ f dµ by definition is ∑n=1∞ ∫En f dµ. Therefore, we can rewrite the equation above as:
lim n→∞ ∫En f dµ = ∑n=1∞ ∫En f dµ.
Combining all the steps, we have:
∫∞ f dµ = ∫X lim n→∞ fn dµ = lim n→∞ ∫X fn dµ = lim n→∞ ∫En fn dµ = lim n→∞ ∫En f dµ = ∑n=1∞ ∫En f dµ.
Thus, we have proven that ∫∞ f dµ = ∑n=1∞ ∫En f dµ using the monotone convergence theorem and linearity of the integral.