Prove that $f(x) - f(a) = \int ^{x}_{a} g(t)dt$ and $f '(x) = g(x)$

Question

Let $I:= [a, b]$ and let $f_{n}$ be a sequence of functions that converge to $f$ on $I$.

Suppose that each derivative $f_n'(x)$ is continuous and the sequence {$f'_n$} converge uniformly to $g$ on $I$

Prove that $f(x) - f(a) = \int ^{x}_{a} g(t)dt$ and $f '(x) = g(x)$

So basically I have no idea how to even begin this proof.

Any hints or insights is deeply appreciated.

Answer 1

You need to show that a uniform limit commutes with taking integrals. With that, $$ \int_a^xg(t)\,dt=\int_a^x\lim_nf'_n(t)\,dt=\lim_n\int_a^xf'_n(t)\,dt =\lim_nf_n(x)-f_n(a)=f(x)-f(a). $$ And, interestingly (and as hinted by the statement) only pointwise convergence $f_n\to f$ is needed.