This question is an offshoot of this one.
In this case, the integral is finite. But if the integral was over an infinite interval, what conditions would need to be added, along with uniform convergence of $f_n \to f$ over this infinite interval, to swap the limit and integral?
What condition to add depends on what background you have.
For convenient let's restrict the domain into the unit open interval $(0,1)$ rather than $\Bbb R$.
I would say the indefinite integral can be seen as a linear operator $\int:L^1(0,1)\to C^0(0,1)/\Bbb R$. If you start with Lebesgue integrable function, the indefinite integral gives you a family of continuous functions which are different by constant functions.
For example $\int x=\frac12x^2+C$. The $\frac12x^2+C=\{\frac12x^2+c:c\in\Bbb R\}$ is a class of continuous functions.
That $\int:L^1(0,1)\to C^0(0,1)/\Bbb R$ is bounded is because the absolute continuous functions are continuous. For the proof you have to read textbooks about real analysis.
So if $f_n\to f$ in $L^1$, which is $\lim_{n\to\infty}\int_0^1|f_n(x)-f(x)|dx=0$ (the true integral on the absolute difference goes to $0$), then $\int f_n\to\int f$ uniformly, in the senses that if we pick $g_n,g\in C^0(0,1)$ such that $\int f_n=g_n+C$ and $\int f=g+C$, then there are numbers $(c_n)_n$ such that $$\lim_{n\to\infty}\sup_{x\in(0,1)}|g_n(x)-g(x)+c_n|=0$$ which is the same as $$\lim_{n\to\infty}\inf_{c\in\Bbb R}\sup_{x\in(0,1)}|g_n(x)-g(x)+c|=0.$$
There are a few well known convergence theorems that outline the different situations where $$\lim_n \int_A f_n(x) = \int_A \lim_n f_n(x).$$ Likely the most often invoked convergence theorem is the dominated convergence theorem: Let $A$ be a measurable subset of $\mathbb{R}$. If $f_n(x) \to f(x)$ for each $x$ (pointwise convergence), and there is some function $g(x)$ with $g(x) \geq |f_n(x)|$ for every $x$ and every $n$, and $\int_A |g(x)| dx < \infty$, then $\lim_n \int_A f_n = \int_A f$.
I am assuming you're not a student of measure theory -- so you may have three questions: (1) What does it mean that A is measurable? (2) What is the purpose of this g function? and (3) how does this answer my original question?
(1). You can think of a measurable subset of $\mathbb{R}$ as a subset. $A$ could be equal to all of $\mathbb{R}$, it could be the positive half of $\mathbb{R}$ or a finite interval, or some isolated points, or some combination of all of the above... The collection of measurable subsets of $\mathbb{R}$ is very large and describing a subset that is not measurable is actually quite hard. This is moving away from your original question so lets leave it at that.
(2) Now what about this bounding function $g$? It is a like a box with finite volume, the two functions $-g$ and $+g$ make up some shape with finite area ($\int|g| < \infty$) that all of the $f_n$ are forced to stay inside of. To understand how this is helpful, its good to look at two situations where you cannot interchange the limit and the integral.
First we have a case where mass escapes to infinity by sliding. Say $f_n(x) = 1 \text{ if } x \in [n,n+1]$ and $f_n(x) = 0$ otherwise. It's pretty clear that $f_n(x) \to 0$ pointwise and also that $\int_\mathbb{R} f_n = 1$ for all $n$. So the limit and the integral cannot be interchanged, we also see that the DCT is not going to work here, there's no finite area shape that will contain all the different $f_n$.
Second we have a case where mass escapes to infinity by "ascending". Say $f(x) = \max(n - |n^2x - n|,0)$. I encourage you to plot these functions for a few values of $n$. These are the so called "marching triangles". We still have that $f_n(x) \to 0$ but $\int_0^1 f_n(x) = 1$ for all $n$.
Now for (3). Notice that in the last two examples one was on a bounded set and the other one was on an unbounded one. The dominated convergence theorem is true regardless of whether the domain of integration is bounded or unbounded. The only difference is that on a bounded domain you have that $\int_A 1 < \infty$ -- so a bound like $|f_n| < C$ will suffice. This bound by a constant prevents mass from escaping upwards, and being on a bounded interval is enough to prevent mass from escaping outwards -- intuitively the two situations where the limit and integral fail to be interchangeable.
