Let $f(x)$ be $$ f(x) = \max\left\{0,x\right\} $$
Which isn't differentiable in $x=0$, In order to approximate such function I'm using the sequence
$$ f_n(x) = \int_{-\infty}^{x} g_n(t)dt=\int_{-\infty}^x \frac{1}{1+e^{-nt}}dt = \frac{1}{n} \int_{-\infty}^{nx} \frac{1}{1+e^{-t}}dt = \frac{1}{n}\left(nx + \log(1+e^{-nx})\right) = \frac{1}{n}\log(1+e^{nx}) $$
And I want to prove that this is correct. In order to prove this I observe that
$$ g_n(x) \leq 1 $$
and that
$$ \lim_{n \to \infty} g_n(x) = u(x) $$
where $u(x)$ is the step function. By the dominate convergence theorem I can then say that
$$ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \int_{-\infty}^{x} g_n(t) dt = \int_{-\infty}^{x} \lim_{n \to \infty} g_n(t) dt = \int_{-\infty}^{x} u(t) dt = \max\left\{0,x\right\} $$
I omitted the measurability of the sequence and $u$, but that's trivial right?
Is everything correct?
Update: I believe that since $$ \forall n \; g_n(x) \leq u(x) $$
the whole argument could work.
Maybe you needed the integrals in your heuristic process leading to $$f_n(x):={1\over n}\bigl( nx+\log(1+e^{-nx})\bigr)\ .$$ But now we can forget about them and just prove that for fixed $x$ one has $$\lim_{n\to\infty} f_n(x)=\max\{0,x\}\ .$$ This is easy. You have to distinguish the cases $x>0$, $x=0$, and $x<0$.