Why a uniform limit of a sequence of bounded derivative is a derivative? Assume that a sequence $(f_n)$ of functions $f_n:[0,1] \rightarrow \mathbb R$ is uniformly convergent to a function $f:[0,1]\rightarrow \mathbb R$. Moreover let (for each $n\in \mathbb N$)  $f_n$ be bounded and $f_n$ be a derivative of some $F_n: [0,1]\rightarrow \mathbb R$.
How to show that $f$ is a derivative of some $F:[0,1]\rightarrow \mathbb R$?
My attempt: by assumptions follows easily that $f_n$ (as derivative)
 is measurable , consequently (as measurable and bounded) $f_n$ is 
integrable. Hence $F_n$
 is absolutely continuous. We have also that 
$f$ is integrable and bounded (as a 
consequence of uniformly convergence). Let 
$$G_n(x)=
\int_0^xf_n(t)dt.$$ 
Then (using definition of uniform convergence)
$$\lim_{n\rightarrow \infty} G_n(x)=
\int_0^x f(t)dt=:G(x) \textrm{ for } x\in [0,1].$$ 
I'm not sure whether 
$G'(x)=f(x)$ for all 
$x∈[0,1]$., because I know only that absolutely continuous function has derivatives  almost everywhere. 
 A: To show that $G'(x) = f(x)$ for an $x \in [0,1]$, we must, for every $\varepsilon > 0$, find (establish the existence of) a $\delta > 0$ such that
$$0 <\lvert h\rvert < \delta \land (x+h)\in[0,1] \Rightarrow \left\lvert \frac{G(x+h)-G(x)}{h} - f(x)\right\rvert < \varepsilon.$$
Let $\varepsilon > 0$ be arbitrary. By the uniform convergence of $f_n \to f$, choose an $N\in \mathbb{N}$ such that $\lvert f_N(y) - f(y)\rvert < \varepsilon/3$ for all $y \in [0,1]$. For that $N$, by the assumed differentiability of $F_N$ in $x$, choose $\delta > 0$ such that
$$0 <\lvert h\rvert < \delta \land (x+h)\in[0,1] \Rightarrow \left\lvert \frac{F_N(x+h)-F_N(x)}{h} - f_N(x)\right\rvert < \varepsilon/3.$$
Then, for admissible $h$, we have
$$\begin{align}
\left\lvert \frac{G(x+h)-G(x)}{h} - f(x)\right\rvert &= \left\lvert \frac1h \int_0^h f(x+t)\,dt - f(x) \right\rvert\\
&= \Biggl\lvert \frac1h \int_0^h \bigl(f(x+t) - f_N(x+t)\bigr)\,dt\\
&\qquad + \left(\frac1h \int_0^h f_N(x+t)\,dt - f_N(x)\right)\\
&\qquad + \bigl(f_N(x) -f(x)\bigr)\Biggr\rvert\\
&\leqslant \frac1{\lvert h\rvert} \int_0^{\lvert h\rvert} \lvert f(x+ \sigma(h)t)-f_N(x+\sigma(h)t)\rvert\,dt\tag{*}\\
&\qquad + \left \lvert\frac{F_N(x+h)-F_N(x)}{h} -f_N(x)\right\rvert\\
&\qquad + \lvert f_N(x) - f(x)\rvert\\
&< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3},
\end{align}$$
so we see that $G$ is differentiable in $x$ with derivative $G'(x) = f(x)$.
In $(*)$, $\sigma(h)$ is the sign of $h$, $\pm 1$.
