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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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    $\begingroup$ 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. $\endgroup$ – davidlowryduda Nov 19 '11 at 19:21
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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 $\int_{\mathbb{N}} b\,d\nu = \sum_{n=1}^\infty a_n$.)

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

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  • $\begingroup$ 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 $\endgroup$ – user3503589 Nov 5 '14 at 9:24
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    $\begingroup$ 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.) $\endgroup$ – Nate Eldredge Nov 5 '14 at 14:48
  • $\begingroup$ 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. $\endgroup$ – user3503589 Nov 6 '14 at 16:54
  • $\begingroup$ Nate: Can you please point out a standard reference for "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 $\int \sum f_n = \sum \int f_n$"? $\endgroup$ – user389066 Aug 15 '17 at 18:43
  • $\begingroup$ @User31443: Are you looking for a reference for the usual general statement of Fubini's theorem (Folland's Real Analysis, for instance) or for how the general statement implies this special case? It is literally just the general case applied with Lebesgue measure and counting measure. $\endgroup$ – Nate Eldredge Aug 15 '17 at 20:38
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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 (non-negative) 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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  • $\begingroup$ Just a little observation: you don't have say that $\phi_j\uparrow f_1$ and $\psi_j\uparrow f_2$. $\endgroup$ – leo Nov 19 '11 at 19:44
  • $\begingroup$ @leo Thanks! I have added this. $\endgroup$ – Jonas Teuwen Nov 19 '11 at 19:46
  • $\begingroup$ No problem Jonas $\endgroup$ – leo Nov 19 '11 at 19:55
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    $\begingroup$ @JonasTeuwen What does $\phi_j \uparrow f_1$ mean? I am not familiar with this notation. $\endgroup$ – Noir Jun 17 '18 at 17:30
  • $\begingroup$ It means that $\{\phi_j\}$ is a monotonically non-decreasing sequence of functions which converges to $f_1$ pointwise. $\endgroup$ – Sean Haight Oct 31 '18 at 15:40
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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])(http://artofproblemsolving.com/community/c7h294262p1593291):

Consider the calculation \begin{align*}\ln 2 &= \int_0^1 \frac1{1+x}\,dx = \int_0^1\sum_{n=0}^{\infty} (-1)^n x^n\,dx\\ ?&= \sum_{n=0}^{\infty}\int_0^1(-1)^n x^n\,dx = 1-\frac12+\frac13-\frac14+\cdots\end{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, $$0\le \sum_{n=0}^{N}(-1)^n x^n\le 1$$ for all $x\in [0,1]$. $1$ is integrable on this interval, and the interchange $$\int_0^1\left(\lim_{N\to\infty}\sum_{n=0}^{N}(-1)^n x^n\right)\,dx = \lim_{N\to\infty}\int_0^1 \sum_{n=0}^{N}(-1)^n x^n\,dx$$ 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.

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