Uniform convergence implies convergence of antiderivative 
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*From this theorem, can we conclude that if $f_n'$ is a sequence of integrable functions that converge uniformly to $f'$, then the sequence of their antiderivatives $f_n$ converges pointwise to the antiderivative $f$ of $f'$?

*If so, when is the convergence uniform as well as pointwise?


I have what I think is a correct proof for the proposition in (1), but in the interest of not distracting from (2) I'll omit it for now.
 A: Your first proposition is not entirely precise (for me), so I'll write what I understand:

  
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*Let $g_n:[0,1]\rightarrow\mathbb{R}$ be a sequence of continuous functions converging uniformly for some function $g:[0,1]\rightarrow$. Then $g$ is continuous and, if $f_n:[0,1]\rightarrow\mathbb{R}$ denote the antiderivative of $g_n$ such that $f_n(0)=0$, then $f_n$ converge pointwise for some continuous function $f:[0,1]\rightarrow\mathbf{R}$ such that $f'=g$ (and $f(0)=0$).
  

Indeed, in this case, each $f_n$ is given by $f_n(x)=\int_0^x g(t)dt$, so by the theorem you mentioned, there exists, for each $x\in[0,1]$, $f(x):=\lim_{n\rightarrow\infty}\int_0^x g_n(t)dt=\int_0^xg(t)dt=\lim_{n\rightarrow\infty}f_n(x)$. Since $f$ is given by the integral of a continuous function, $f$ is derivable and $f'=g$.

$\ $$\ $ $\text{2}$. In the case above, the convergence of $f$ is always uniform.

Note that, for every $x\in [0,1]$,
$$|f(x)-f_n(x)|=|\int_{0}^x g(x)-g_n(x)dx\leq\int_0^1|g(x)-g_n(x)|\leq\Vert{g-g_n}\Vert_\infty$$
Since the convergence of the $g_n$'s is uniform, so is the $f_n's$.
