Interchanging integral and sum in case of absolute and uniform convergence Suppose $(X, \mathcal{X}, \mu)$ is a measure space, $\{f_n\}_{n \ge 1} \subseteq L^1(X,\mathcal{X}, \mu)$ is a sequence of real-valued functions on $X$, and $f \colon X \to \mathbb{R}$ is a function such that
$$
f(x) = \sum_{n \ge 1} f_n(x), \quad x \in X
$$
where the convergence is absolute and uniform.
When can I say that
$$
\int_X \sum_{n \ge 1} f_n \, \mathrm{d}\mu = \sum_{n \ge 1} \int_X f_n \, \mathrm{d}\mu \;? \tag{1}
$$
I know that to be able to say that, a sufficient condition would be
$$
\int_X \sum_{n \ge 1} |f_n| \, \mathrm{d}\mu < \infty
$$
(or equivalently $\sum_{n \ge 1} \int_X |f_n| \, \mathrm{d}\mu < \infty$) by Fubini's theorem.
Now, absolute convergence implies that for $x \in X$, $g(x) := \sum_{n \ge 1} |f_n(x)| < \infty$, and uniform convergence implies that for all $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for every $n \ge N$ we have $\sup_{x \in X} |f(x) - \sum_{i=1}^n f_i(x)| < \varepsilon$.
I am unable to show (1) with just these assumptions. Is it even possible? If not what other regularity conditions can I impose on $f$ to be able to say that.
 A: In order to show (1) an assumption like uniform convergence is often not very helpful. If your measure space $X$ has infinite measure then uniform convergence doesn't imply convergence in $L^1$. To see an example of this first note that your question is equivalent to the same question but where you consider sequences of functions converging to a given function in stead of sums of functions converging to a given function. Consider the sequence $f_n(x) = \frac{1}{n}\mathbb{1}_{[n^2, n^2+n]}$. This is a sequence of $L^1((0, \infty))$ functions(where we are considering the usual Lebesgue measure) that converges uniformly to the $0$ function but
$$
1 = \lim_{n \to \infty}\int f_n \neq \int \lim_{n \to \infty}f_n = \int 0 = 0
$$
In order to do the type of thing you are talking about one usually uses a theorem called the Lebesgue dominated convergence theorem. This theorem gives sufficient but not necessary conditions for switching a limit and and an integral. In some cases you might want to look at the Vitali convergence theorem. This theorem gives necessary and sufficient conditions for switching a limit and an integral but it is more complicated and harder to work with than the Lebesgue dominated convergence theorem.
