How do I prove that if $f$ is continuous, then $F_{0}'(x) = f(x)$? If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$. 
Define a function $F_0(x) = \int_a^x f$ for $a \le x \le b$
If $x$ and $x+h$ both lie in $I$, then 
$F_{0}(x+h) -F_{0}(x) = \int_a^{x+h} f - \int_a^x f$
= $\int_x^{x+h} f = f(\bar{x})h$
where we use the mean value theorem for integrals and $\bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $\bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)
Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$? 

Following your argument, you have that $$\frac{F_0(x_0+h)-F_0(x_0)}{h}=f(\bar x),$$ but the mean value theorem aserts that $\bar x\in[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $\bar x$ depends on $h$.
So if $h\to0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $\bar x\to x_0$ as $h\to 0$. So, if you put all together, you'll have $$F_0'(x_0)=\lim_{h\to0}\frac{F_0(x_0+h)-F_0(x_0)}{h}=\lim_{h\to0}f(\bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.
