differentiability of $\int_{a}^{x}f$ at a jump discontinuity point of $f$. Let $f:[a,b]\rightarrow\mathbb{R}$ be an integrable function and let $F(x)=\int_{a}^{x}f$. if $c\in(a,b)$ is a jump discontinuity point of $f$ then $F$ is not differentiable at $c$.
I don't know how to show that that limit
$$\lim_{h\rightarrow0} \frac{F(c+h)-F(c)}{h}$$ doesn't exist but I managed to show that it can't be equal to $f(c)$. Maby I can use the fundamental theorem of calculus somehow?
Can someone please help me and check my work so far?
My work:
$$\frac{F(c+h)-F(c)}{h} = \frac{1}{h}\cdot\left(\int_{a}^{c+h}f-\int_{a}^{c}f\right)
=\frac{1}{h}\int_{c}^{c+h}f$$
$f$ is not continuous at $c$ so there is an $\epsilon>0$ such that every $\delta>0$
if $x\in(c-\delta, c+\delta)$ then $f(c)+\epsilon\leq f(x)\leq f(c)-\epsilon$
Let $P=\{x_0,...,x_n\}$ be a partition of $[c,c+h]$.
$$\int_{c}^{c+h}f = \lim_{n\rightarrow\infty} \sum\limits_{i=1}^{n} m_i\cdot\Delta x_i$$ (Darboux integral)
Where $m_i=\inf\{f(x):x\in[x_{i-1},x_i]\}$ and $\Delta x_i=x_i-x_{i-1}$
$f(x)\leq f(c)-\epsilon$ so $m_i\leq f(c)-\epsilon$.
We got that
$$\int_{c}^{c+h}f = \lim_{n\rightarrow\infty} \sum\limits_{i=1}^{n} m_i\cdot\Delta x_i\leq
\lim_{n\rightarrow\infty}\sum\limits_{i=1}^{n}(f(c)-\epsilon)\Delta x_i =
\lim_{n\rightarrow\infty}(f(c)-\epsilon)\sum\limits_{i=1}^{n}\Delta x_i =
(f(c)-\epsilon)\cdot (c+h-c)$$
So
$$\frac{1}{h}\int_{c}^{c+h}f\leq(f(c)-\epsilon)$$
Hence
$$\lim_{h\rightarrow0} \frac{F(c+h)-F(c)}{h} \neq f(c)$$
 A: Here is my solution:
We want to show that the limit:
$$\lim_{h\to 0}\frac{F(c+h)-F(c)}{h}$$ doesn't exist.
I'll do so by showing that the left- and right- hand limits are not equal to each other.
$$\frac{F(c+h)-F(c)}{h}=\frac{\int_{a}^{c+h}f-\int_{a}^{c}f}{h}=\frac{1}{h}\int_{c}^{c+h}f$$
Let $L=\lim_{x\to c^{-}} f(x)$, R=$\lim_{x\to c^{+}} f(x)$
$c$ is a jump discontinuity point of $f$ so $L\neq R$
Let $P=\{x_0,...x_n\}$ be a partition of $[c,c+h]$ and $(\xi_i)_{i=1}^{n}$ is a sequence of points such that $\xi_i\in[x_{i-1},x_i]$ thus
$$\int_{c}^{c+h} f=\lim_{\lambda(p)\rightarrow0} \sum\limits_{i=1}^{n} f(\xi_i)\Delta x_i$$  (Riemann integral)
Suppose $h\rightarrow 0^{+}$
for every $\epsilon>0$ there is a $\delta>0$ such that if $c\leq x\leq c+ \delta$ then$|f(x)-R|<\epsilon$
so if $h<\delta$, for every $c\leq x\leq c + h$ $\Rightarrow$ $|f(x)-R|\leq\epsilon$
By the definition of $(\xi_i)$ we know that $c\leq\xi_i\leq c+h$ for every i, so $|f(\xi_i)-R|\leq\epsilon$
Because $\xi_i$ is an arbitrary point in $[x_{i-1}, x_i]$ we can define it such that
$f(\xi_i)\leq R+\epsilon$
so
$$\int_{c}^{c+h} f=\lim_{\lambda(p)\rightarrow0} \sum\limits_{i=1}^{n} f(\xi_i)\Delta x_i\leq\lim_{\lambda(p)\rightarrow 0}\sum\limits_{i=1}^{n} (R+\epsilon)\Delta x_i =
(R+\epsilon)\cdot h$$
therefore
$$\frac{1}{h}\int_{c}^{c+h}f\leq R+\epsilon$$
hence
$$\lim_{h\to 0^{+}}\frac{F(c+h)-F(c)}{h}=\lim_{h\to 0^{+}}\frac{1}{h}\int_{c}^{c+h}f=  R$$
Similarly we can show that
$$\lim_{h\to 0^{-}}\frac{F(c+h)-F(c)}{h}=\lim_{h\to 0^{-}}\frac{1}{h}\int_{c}^{c+h}f=  L$$
We know that $R\neq L$  so
$$R=\lim_{h\to 0^{+}}\frac{F(c+h)-F(c)}{h} \neq \lim_{h\to 0^{-}}\frac{F(c+h)-F(c)}{h}=L$$
Thus the limit $$\lim_{h\to 0}\frac{F(c+h)-F(c)}{h}$$ doesn't exist and F is not differentiable at $c$
