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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

2 Answers 2

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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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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