If $f$ is absolutely continuous and $g$ is continuous, prove $f' =g$.

Question

The full problem reads:

Prove that if $f$ is absolutely continuous on $[0,1]$ and $g$ is continuous on $[0,1]$ such that $f'=g$ a.e., then $f$ is differentiable on $[0,1]$ and $f'=g$.

My analysis skills are very rusty and I'm having a hard time seeing how to prove this. Thanks in advance for any advice!

Answer 1

By Lebesgue's fundamental theorem of calculus, $$f(x)=f(0)+\int_0^xf'. $$ By hypothesis, $$f(x)=f(0)+\int_0^x g.$$ By the standard fundamental theorem of calculus, $f'(x)=g(x)$ for all $x$.