Let $f$ be a real-valued Riemann integrable function on $[0,1]$. Let $F(x)=\int_0^x f(t)dt$.
Prove that $F'(x)=f(x)$, for almost all $x\in[0,1]$.
If it's continuous, then certainly Riemann so linearity holds (I can move things in). Since continuous, I can use the definition ($\epsilon,\delta$ definition) to prove the statement is true.
From the problem, it says "for almost all", so I guess I need to consider Lebesgue. Since Riemann then certainly Lebesgue. However, I'm still stuck...
I'm not sure if this is an answer or a comment. By Lebesgue's biconditional criterion for Riemann integrability, a function is Riemann integrable iff it is bounded and continuous almost everywhere. By the Fundamental Theorem of Calculus, the function $F$ is differentiable at each point where $f$ is continuous, and at these points, $F'(x)=f(x)$.