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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 $x$, $F'(x) = f(x).$

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  • $\begingroup$ to be or to be or not to be, that is the question $\endgroup$
    – Hello
    Nov 28, 2016 at 22:01

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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?)

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  • $\begingroup$ Looking at the definition of differentiability and after graphing your function it looks like the limit from the left and right side at any $x\in[0,1]$ is equal, so I am not sure. $\endgroup$
    – Remy
    Nov 28, 2016 at 19:02
  • $\begingroup$ Or would the limit at x=$1\over{2}$ not exist? I don't know how that works for isolated points. $\endgroup$
    – Remy
    Nov 28, 2016 at 19:05
  • $\begingroup$ Well, the function $\;f\;$ is clearly discontinuous, but that's not the point: by Darboux Theorem, the derivative of a function always fulfills the intermedaite value property, so $\;f= F'\;$ for some $\;F\;$ , as $\;F'(0)=0,\,F'(0.5)=1\;$ then $\;F'\;$ should attain all the intermediate values... $\endgroup$
    – DonAntonio
    Nov 28, 2016 at 19:10
  • $\begingroup$ Ok so it can be discontinuous but any F'(x) between 0 and 1 must be defined? $\endgroup$
    – Remy
    Nov 28, 2016 at 19:14
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    $\begingroup$ @JohnH Indeed so. $\endgroup$
    – DonAntonio
    Nov 28, 2016 at 19:30

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