Suppose $f$ has a discontinuity at $c \in (a,b)$, show that $F(x) = \int_a^x f$ is not differentiable at $c$ 
Suppose $f : [a,b] \to \mathbb{R}$ is increasing, by Proposition 5.2.11, $f$ is Riemann integrable. Suppose $f$ has a discontinuity at $c \in (a,b)$, show that $F(x) = \int_a^x f$ is not differentiable at $c$.
Proposition 5.2.11. Let $f: [a,b] \to \mathbb{R}$ be a monotone function. Then $f$ is Riemann integrable.

I know that the second FTC says that if $f$ is continuous at $c$, then $F$ is differentiable at $c$. This question is to prove the reverse. I know that without assuming that $f$ is increasing, this does not hold. The textbook provides the counter-example. $f(x) = 0$ if $x \not=0$, $f(0) =1$, and let $F(x) = \int_0^x f$. In this case, $f$ is not continuous at $0$, but $F'(0)$ exists. So, I think that I should use some property of an increasing function, but I can't figure it out. I appreciate if you give some help.
 A: Since $f$ is increasing, both limits$$l=\lim_{x\to c^-}f(x)\text{ and }L=\lim_{x\to c^+}f(x)$$must exist. And, since $f$ is discontinuous at $c$, $l\neq L$. Then, clearly, $l<L$.
Consider the function $G(x)=\int_c^xf(t)\,\mathrm dt$. Since $G-F$ is constant, $F$ is differentiable at $c$ if and only if $G$ is differentiable at $c$. But, for each $t>c$, $f(t)>L$ and therefore$$\frac{G(x)-G(c)}{x-c}=\frac{\int_c^xf(t)\,\mathrm dt}{x-c}\geqslant L.$$So, the right derivative of $G$ at $c$, if it exists, is greater than or equal to $L$. And, by the same argument, the left derivative of $G$ at $c$, if it exists, is smaller than or equal to $l$. In any case, $G$ is not differentiable at $c$.
A: Assume the function is monotone increasing (the decreasing case is similar). Then we get for $h>0$
$$ \frac{1}{h}(F(c+h) -F(c))= \frac{1}{h} \int_c^{c+h} f(x)dx \geq f(c^+) $$
and for $h<0$
$$ \frac{1}{h}(F(c+h) -F(c))=\frac{1}{h} \int_c^{c+h} f(x)dx \leq f(c^-). $$
Assume that $F$ is differentiable in $c$, then we get (by taking the limit $h\rightarrow 0$)
$$ f(c^+) \leq F'(c) \leq f(c^-). $$
However, $f$ is discontinuous at $c$ and monotone increasing, so we have $f(c^+) < f(c^-)$ which yields a contradiction. 
