Existence of Differentiable Function 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.
Your question has a negative answer. For instance the function defined by $f(x)=\cases{-1,&amp; $-1\le x&lt;0$ \cr 1,&amp;$ 0\le x\le 1$}$ is Riemann integrable over $[-1,1]$, but is not the derivative of any function (since derivatives have the intermediate value property as a result of Darboux's Theorem).
Note that, A bounded function $f$ of $[a,b]$ is Riemann integrable on $[a,b]$ iff the function is continuous almost everywhere. This is known as Lebesgue's criteria.