Dtyjlui

Let's say I have $int_{0}^{infty}sum_{n = 0}^{infty} f_{n}(x), dx$ with $f_{n}(x)$ being continuous functions. When can interchange the integral and summation? Is $f_{n}(x) geq 0$ for all $x$ and for all $n$ sufficient? How about when $sum f_{n}(x)$ converges absolutely? If so why?

I like to remember this as a special case of the Fubini/Tonelli theorems, where the measures are counting measure on $mathbb{N}$ and Lebesgue measure on $mathbb{R}$ (or $[0,infty)$ as you've written it here). In particular, Tonelli's theorem says if $f_n(x) ge 0$ for all $n,x$, then $$sum int f_n(x) dx = int sum f_n(x) dx$$ without any further conditions needed. (You can also prove this with the monotone convergence theorem.)

Then Fubini's theorem says that for general $f_n$, if $int sum |f_n| < infty$ or $sum int |f_n| < infty$ (by Tonelli the two conditions are equivalent), then $int sum f_n = sum int f_n$. (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 $(X,mathcal{F}, mu),(Y,mathcal{G}, nu)$ be $sigma$-finite measure spaces, and let $g : X times Y to mathbb{R}$ be measurable with respect to the product $sigma$-algebra $mathcal{F} otimes mathcal{G}$. Suppose that $int_X int_Y |g(x,y)| nu(dy) mu(dx)$ is finite. (Note: By Tonelli's theorem, this happens if and only if $int_Y int_X |g(x,y)|mu(dx)nu(dy)$ is finite, since both iterated integrals are equal.) Then $$int_X int_Y g(x,y) nu(dy)mu(dx) = int_Y int_X g(x,y) mu(dx) nu(dy).$$

Let $X = mathbb{R}$, $mathcal{F}$ the Borel $sigma$-algebra, and $mu$ Lebesgue measure. Let $Y = mathbb{N}$, $mathcal{G} = 2^{mathbb{N}}$ the discrete $sigma$-algebra, and $nu$ counting measure. Define $g(x,n) = f_n(x)$. Exercise: since each $f_n$ is measurable, verify that $g$ is measurable with respect to $mathcal{F} otimes mathcal{G}$. 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 $a_n$, define a function $b : mathbb{N} to mathbb{R}$ by $b(n) = a_n$. Then $int_{mathbb{N}} b,dnu = sum_{n=1}^infty a_n$.)

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

This is a theorem that will work:

Theorem. If ${f_n}_n$ is a positive sequence of integrable functions and $f = sum_n f_n$ then
$$int f = sum_n int f_n.$$

Proof. Consider first two functions, $f_1$ and $f_2$. We can now find sequences ${phi_j}_j$ and ${psi_j}_j$ of (nonnegative) simple functions by a basic theorem from measure theory that increase to $f_1$ and $f_2$ respectively. Obviously $phi_j + psi_j uparrow f_1 + f_2$. We can do the same for any finite sum.

Note that $int sum_1^N f_n = sum_1^N int f_n$ for any finite $N$. Now using the monotone convergence theorem we get

$$sum int f_n = int f.$$

Note 1: If you're talking about positive functions absolute convergence is the same as normal convergence as $|f_n| = f_n$.

Note 2: Continuous functions will be certainly integrable if they have compact support or tend to $0$ fast enough as $x to pm infty$.

While most of the time I would use the Fubini/Tonelli conditions, the dominated convergence theorem is actually strictly stronger in this mixed sum/integral case, because it can take into account the order structure of the integers. An example (that I first worked up back in 2009:

Consider the calculation
begin{align*}ln 2 &= int_0^1 frac1{1+x},dx = int_0^1sum_{n=0}^{infty} (-1)^n x^n,dx\
?&= sum_{n=0}^{infty}int_0^1(-1)^n x^n,dx = 1-frac12+frac13-frac14+cdotsend{align*}
Fubini's theorem isn't strong enough to justify the interchange. If we put absolute values on the terms, it blows up to $int_0^1 frac1{1-x},dx = 1+frac12+frac13+frac14+cdots=infty$.

On the other hand, the dominated convergence theorem cares about the partial sums $sum_{n=0}^{N}(-1)^n x^n$. By the alternating series estimate,
$$0le sum_{n=0}^{N}(-1)^n x^nle 1$$
for all $xin [0,1]$. $1$ is integrable on this interval, and the interchange
$$int_0^1left(lim_{Ntoinfty}sum_{n=0}^{N}(-1)^n x^nright),dx = lim_{Ntoinfty}int_0^1 sum_{n=0}^{N}(-1)^n x^n,dx$$
is justified, proving the result $1-frac12+frac13-frac14+cdots=ln 2$.

This situation with the dominated convergence theorem being stronger than Fubini's theorem can come up when we've got a reasonable bound on partial sums but not absolute convergence as a whole.

The monotone convergence theorem, on the other hand, is exactly the same as Tonelli's theorem - when everything's positive, either both sides are the same and finite or both sides are infinite.