I'd like a confirm that my solution is correct, thank you in advance.
Suppose $(X, B, \mu)$ is a measure space, $(f_n)_n \subset L^1(\mu)$ and $f_n \to f$ uniformly for some $f:X \to \mathbb{R}$.
a) If $\mu(X) < \infty$, then $f \in L^1$ and $\int f_n \to \int f$.
b) If $\mu(X) = \infty$, the conclusion of a) can fail.
a) Since $f_n \to f$ pointwise $f$ is measurable and we have $\int |f(x)|d\mu \le \int(||f_n - f||_{\infty}+|f_n(x)|)d\mu=||f_n - f||_{\infty}\mu(X) + \int |f_n(x)|d\mu < \infty$ if $n$ is big enough. Therefore $f \in L^1$.
Now $|f_n(x)|\le ||f_n - f||_{\infty} +|f(x)|\le 1+|f(x)|$ for $n$ big enough (say $n\ge N$) and the function on the right side is $L^1$ since $\mu$ is finite.
So $(f_n)_n$ is dominated by an $L^1$ function for $n\ge N$ and converges pointwise to $f$ (which is measurable), by the dominated convergence theorem we obtain that $\int f_n \to \int f$.
b) Let $(X, B, \mu)=(\mathbb{R}, \mathcal{L}, m)$ and consider $f_n(x)=\frac{1}{n}1_{[-n,n]}(x)$, $f=0$.
We have $(f_n)_n \subset L^1$, since $\int f_n =2$ for every $n$, and $||f_n -f||_{\infty}=1/n \to 0$, but $\lim_n \int f_n = 2 \ne 0 = \int f$.