If $f_n \rightrightarrows f$ and every $f_n$ has antiderivative is that true that $f$ has antiderivative? There is a sequence of functions $f_n$ and each of these functions has antiderivative. If they are uniformly convergent to $f$ is that true that $f$ has antiderivative?
I know that is true in case of continuity and that every continuous function has antiderivative but on the other hand, I know that there are functions that are not continuous but have antiderivatives.
 A: The answer is yes. The following is a classic theorem (see for example Baby Rudin somewhere in the chapter on uniform convergence):

Let $\{\phi_n\}$ be a sequence of differentiable functions $[a,b]\to\Bbb{R}$ such that $\phi_n'$ converges uniformly to some function $\psi:[a,b]\to\Bbb{R}$, and such that for some $x_0\in [a,b]$, we have that $\{\phi_n(x_0)\}_{n=1}^{\infty}$ converges. Then, $\{\phi_n\}$ converges uniformly to a function $\phi$ on $[a,b]$, and $\phi'=\psi$.

So, in your situation, let $F_n$ be an antiderivative of $f_n$ (which exists by assumption), and consider $\phi_n= F_n- F_n(0)$. In this way, $\phi_n$ is an antiderivative of $f_n$, $\phi_n(0)=0$, so $\lim\limits_{n\to\infty}\phi_n(0)$ exists, and by assumption, $\phi_n'=f_n$ converges uniformly. Thus, the theorem above applies and tells us that $\phi:=\lim\limits_{n\to\infty}\phi_n$ satisfies $\phi'=f$.
Note that there's nothing special about $0$ here. If you're working on an interval $[a,b]$, then again, fix a point $x_0\in [a,b]$, consider an antiderivative $F_n$ of $f_n$, and let $\phi_n=F_n-F_n(x_0)$. All we're doing here is offsetting the constant to ensure pointwise convergence at a single point so that we can invoke the theorem above.
