Fundamental theorem of calculus left limit

Question

Suppose $f(x)$ is integrable and $F(x) = \int_{a}^{x}f(t)dt$. How to show that if $f(x)$ has left limit at $x_0$ then $F'_{-}(x_0) = f(x_0^{-})$.

Answer 1

$$\left|\frac{F(x_0+h)-F(x_0)}{h}-f(x_0^-)\right|\leq \frac{1}{h}\int_{x_0}^{x_0+h}|f(t)-f(x_0^-)|dt.$$ Let $\varepsilon>0$. Since $f$ has a limit at $x_0^-$, there is $\tilde h>0$ small enough s.t. $$|f(t)-f(x_0^-)|<\varepsilon,$$ for all $t\in ]x_0-\tilde h,x_0[.$ In particular, if $0<h<\tilde h$, $$\frac{1}{h}\int_{x_0}^{x_0+h}|f(t)-f(x_0^-)|dt\leq \varepsilon.$$ The claim follow.