When can a sum and integral be interchanged?

Question

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 we 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?

Answer 1

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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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.

Answer 2

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 $x \to \pm \infty$.

Answer 3

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.