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.

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.

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.

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.