Let $f : [a, b] → \mathbb{R}$. Suppose that $f$ is Riemann integrable on $[a, b]$.
Define $F : [a, b] → \mathbb{R}$ by $F(x) := \int_{a}^{x}f(t) dt.$
Then $F$ is differentiable and for each $x$, $F'(x) = f(x).$”
Show that the above statement is true or not.
My start at a solution: Since $f : [a,b] \rightarrow \mathbb{R}$ is Riemann integrable, it follows that its upper and lower integrable are equal to each other. To show that $F$ is differentiable, we must show that for each $c\in[a,b],$ $\lim_{h\to0^-}$$F(c+h)-F(c)\over{h}$ = $\lim_{h\to0^+}$$F(c+h)-F(c)\over{h}$. Without loss of generality, we may assume $c+h\in[a,b]$ I don't know how to prove this is true and am not sure where to start with showing that for each $x$, $F'(x) = f(x).$
It is false without the requirement that $\;f\;$ is continuous. For example:
$$f(x)=\begin{cases}1,&x=\frac12\\{}\\0,&0\le x\le1 ,\,\,x\neq\frac12\end{cases}$$
Then $\;f\;$ is integrable (Riemann or whatever) in $\;[0,1]\;$, yet it can not be the derivative of another function (why?)