Let $f: [a,b] \longrightarrow \mathbb{R}$ integrable and right continuous in $x_{0} \in [ab)$. Prove that $F:[a,b] \longrightarrow \mathbb{R}$, defined by $F(x) = \int\limits_{a}^{x}f(t)\mathrm{d}t$, is right differentiable in $x_{0}$, with $F'_{+}(x_{0}) = f(x_{0})$.
$\textbf{My idea:}$ Calculate $\displaystyle \lim_{x\to x_{0}+}\frac{F(x)-F(x_{0})}{x-x_{0}}$. Following, $\displaystyle \lim_{x\to x_{0}+}\frac{F(x)-F(x_{0})}{x-x_{0}} = \lim_{x\to x_{0}+}\frac{\int\limits_{x_{0}}^{x}f(t)\mathrm{d}t}{x-x_{0}}$. By Mean Value Theorem for Integrals, theres exist $c_{x} \in [x_{0},x]$ such that $\int\limits_{x_{0}}^{x}f(t)\mathrm{d}t = f(c_{x})(x - x_{0})$. Therefore, $\displaystyle \lim_{x\to x_{0}+}\frac{F(x)-F(x_{0})}{x-x_{0}} = \lim_{x\to x_{0}+}f(c_{x}) = f(x_{0})$, since $f$ is right continuous in $x_{0}$.
Is the correct idea?
Not really because for Mean Value Theorem for Integrals, one needs the condition that $f$ being continuous on the whole $[x_{0},x]$, not only right continuous at $x_{0}$.
But we can choose for $\epsilon>0$ a $\delta>0$ such that $0\leq x-x_{0}<\delta$, then $|f(x)-f(x_{0})|<\epsilon$, then \begin{align*} \left|\dfrac{1}{x-x_{0}}\int_{x_{0}}^{x}f(t)dt-f(x_{0})\right|&=\left|\dfrac{1}{x-x_{0}}\int_{x_{0}}^{x}(f(t)-f(x_{0}))dt\right|\\ &\leq\dfrac{1}{x-x_{0}}\int_{x_{0}}^{x}|f(t)-f(x_{0})|dt\\ &\leq\dfrac{1}{x-x_{0}}\int_{x_{0}}^{x}\epsilon dt\\ &=\epsilon. \end{align*}