Suppose $f_1, f_2, . . .$ are (Riemann) integrable functions. Then what is the $\epsilon$ definition of $$\lim_{n \rightarrow \infty} \lim_{M \rightarrow \infty} \int_{0}^{M} f_n(x) dx = L $$ for $L \in \mathbb{R}$?
Does this mean you fix an $N$, take the inner limit, then take the outer limit? I realized I didn't know what this meant when I was trying to show that if $f_n$ converges uniformly then $$\int_{-\infty}^{\infty} |f_n(x)| dx = \int_{-\infty}^{\infty} |f(x)| dx$$ (which I suspect it's true, but I'm not sure if it is).
Note that $a_n = \displaystyle\lim_{M \rightarrow \infty} \int_{0}^{M} f_n(x) dx$ is a sequence. So "$\displaystyle\lim_{n \rightarrow \infty} \lim_{M \rightarrow \infty} \int_{0}^{M} f_n(x) dx = L$" means that
"For all $\epsilon > 0$ there exists an $N$ such that for all $n > N$ we have $\left|\left[\displaystyle\lim_{M \rightarrow \infty} \int_{0}^{M} f_n(x) dx\right] - L\right| < \epsilon$."
As for your conjecture that $\displaystyle\int_{-\infty}^{\infty}|f_n(x)|\,dx = \int_{-\infty}^{\infty}|f(x)|\,dx$, notice that the left side can depend on $n$ but the right side can not depend on $n$. So, there will be many examples where this isn't true.
You might have meant "If $f_n$ converges uniformly to $f$, $\displaystyle\lim_{n \to \infty}\int_{-\infty}^{\infty}|f_n(x)|\,dx = \int_{-\infty}^{\infty}|f(x)|\,dx$."
Sadly, this isn't true either. Consider $f_n(x) = \begin{cases}\frac{1}{n}\left(1-\frac{|x|}{n}\right) & |x| \le n \\ 0 & |x| > n\end{cases}$.
Clearly, $|f_n(x)| \le \frac{1}{n}$ for all $x \in \mathbb{R}$. Hence $f_n$ converges uniformly to $f = 0$.
However, $\displaystyle\int_{-\infty}^{\infty}|f_n(x)|\,dx = 1$ for all $n$ whereas $\displaystyle\int_{-\infty}^{\infty}|f(x)|\,dx = 0$.