Q: If $f$ is Riemann integrable over $[0,1]$, then there exists a differentiable function $F$ on $[a, b]$ such that $F'=f$ and $\int\_a^b f(x)dx = F(b)-F(a)$.

Here's what I'm thinking so far:

* If $\alpha = x$, then you can say that $f$ is also Riemann-Stieltjes integrable on $[0,1]$.
* Then, since $f$ must be a bounded real-valued function due to being Riemann integrable, we can use the theorem that states "For $0\le x \le 1$, put $F(x) = \int\_0^x f(t) dt$. Then $f$ is continuous on $[0,1]$
* In order for $F$ to be differentiable, $f$ must be continuous on $[0,1]$, so we add that as an assumption
* Then, by Fundamental Theorem of Calculus, we can say $\int\_0^1 f(x) \; dx = F(1) - F(0)$.

I realize there are lots of issues here, but I'm hoping I'm at least on the right track. Any direction or hints are appreciated.