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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?

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I'm used to proving it capable with monotone convergence or the Lebesgue dominated convergence methods. But those are hardly sharp, I think. There are many versions of MC and LDC, so I don't know which you know. – mixedmath Nov 19 '11 at 19:21

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.

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I am a bit confused because first you say that $\sum \int f_n(x) dx=\int \sum f_n(x) dx$ holds if $f_n(x) \geq 0$which implies that even if the double integral is not finite the equality does hold and then you quote the Fubini theorem which says that the two integrals are equal if the double integral is finite. I would be grateful if you could explain this to me? Its probably as very stupid question but it makes me very uncomfortable – user3503589 Nov 5 '14 at 9:24
Yes, that's what I said - I'm not sure what part is confusing you? Notice that I did not say "if and only if"! The two theorems give two different hypotheses, each of which leads to the same conclusion. (And note carefully the appearance of the absolute value bars in the hypothesis of Fubini's theorem.) – Nate Eldredge Nov 5 '14 at 14:48
Ah that was my confusion. I was thinking of the case when $f_n(x) \geq 0$ and $\int \sum f_n(x) \to \infty$ and therefore Fubini theorem would be invalid(but luckily as you mentioned, its not"if and only if". Thank you for clearing it up Nate, though it was a stupid question. – user3503589 Nov 6 '14 at 16:54

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 $x \to \pm \infty$.

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Just a little observation: you don't have say that $\phi_j\uparrow f_1$ and $\psi_j\uparrow f_2$. – leo Nov 19 '11 at 19:44
@leo Thanks! I have added this. – Jonas Teuwen Nov 19 '11 at 19:46
No problem Jonas – leo Nov 19 '11 at 19:55

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