If $f\in \mathcal R(I)$ ,$F(x)=\int_{a}^{x}f(t)dt$ differentiable on $I$,then $f\in \mathcal C(I)?$ Show that 
if a bounded function $f$ is a  monotonic the function  on a closed interval $[a,b]$ and the function $F(x)=\int_{a}^{x}f(t)dt $ is differentiable on $[a,b]$,then $f$ is continuous on $[a,b].$
I want to proof it by contradiction applying Darboux theorem on $F(x)=\int_{a}^{x}f(t)dt $.I tried, but it seemingly not works.Do you have some hints? Any of your help will be appreciated!
 A: Suppose $x_0$ is a (jump) discontinuity of $f$ in $I$. This must correspond to a non-existent derivative of $F$ at $x_0$, indeed (suppose it is discontinuous from the right): $f(x_0-\delta) \leq f(x_0)$ and also $f(x_0) < \lambda + f(x_0+\delta)$ for a positive $\lambda$ and for all $\delta>0$.
Now, partition  $[x_0 - \delta, x_0]$ into $\{\Delta x_i \}$ and evaluation points $\{ t_i \}$, and partition [x_0, x_0 + \delta] to $\{\Delta X_i \}$ and $\{T_i \}$. $f\in \mathcal{R}(I)$ and so the riemann sums are guaranteed to converge. We take $\Delta X_i =\Delta x_i$ for all $i\in\{1,\dots,n\}$.
$$I_\mathrm{L}=\int_{x_0 -\delta}^{x_0} f(x)\mathrm{d}x = \lim_n\sum_{i=1}^n f(t_i)\Delta x_i < \lim_n \sum_{i=1}^n(\lambda + f(T_i)) \Delta X_i$$
The right summand is actually $\int_{x_0}^{x_0+\delta}(\lambda+f(x))\mathrm{d}x = \lambda \cdot \delta + I_\mathrm{R}$.
Back to our problem:
$$\lim_{\delta\rightarrow 0} \frac{F(x_0)-F(x_0-\delta)}{\delta} = \lim_{\delta\rightarrow 0}\frac{1}{\delta} I_\mathrm{L}<\lambda + \lim_{\delta\rightarrow 0}\frac{1}{\delta} I_\mathrm{R}=\lambda + \lim_{\delta\rightarrow 0} \frac{F(x_0+\delta)-F(x_0)}{\delta}$$ And this contradicts the assumption that $F'(x_0)$ exists.
