Let’s say I have ∫∞0∑∞n=0fn(x)dx with fn(x) being continuous functions. When can we interchange the integral and summation? Is fn(x)≥0 for all x and for all n sufficient? How about when ∑fn(x) converges absolutely? If so why?

**Answer**

I like to remember this as a special case of the Fubini/Tonelli theorems, where the measures are counting measure on N and Lebesgue measure on R (or [0,∞) as you’ve written it here). In particular, Tonelli’s theorem says if fn(x)≥0 for all n,x, then ∑∫fn(x)dx=∫∑fn(x)dx without any further conditions needed. (You can also prove this with the monotone convergence theorem.)

Then Fubini’s theorem says that for general fn, if ∫∑|fn|<∞ or ∑∫|fn|<∞ (by Tonelli the two conditions are equivalent), then ∫∑fn=∑∫fn. (You can also prove this with the dominated convergence theorem.)

There may be weaker conditions that would also suffice, but these tend to work in 99% of cases.

Elaborating on request: the usual statement of Fubini's theorem goes something like this:

Let (X,F,μ),(Y,G,ν) be σ-finite measure spaces, and let g:X×Y→R be measurable with respect to the product σ-algebra F⊗G. Suppose that ∫X∫Y|g(x,y)|ν(dy)μ(dx) is finite. (Note: By Tonelli's theorem, this happens if and only if ∫Y∫X|g(x,y)|μ(dx)ν(dy) is finite, since both iterated integrals are equal.) Then ∫X∫Yg(x,y)ν(dy)μ(dx)=∫Y∫Xg(x,y)μ(dx)ν(dy).

Let X=R, F the Borel σ-algebra, and μ Lebesgue measure. Let Y=N, G=2N the discrete σ-algebra, and ν counting measure. Define g(x,n)=fn(x). Exercise: since each fn is measurable, verify that g is measurable with respect to F⊗G. Exercise: verify that integration with respect to counting measure is the same as summation, where the integral exists and is finite iff the sum converges absolutely. (That is, given a sequence of real numbers an, define a function b:N→R by b(n)=an. Then ∫Nbdν=∑∞n=1an.)

As such, the conclusion of Fubini's theorem reduces to the statement that was to be proved.

**Attribution***Source : Link , Question Author : user192837 , Answer Author : Michael Hardy*