Let $f: \ [a, b]\rightarrow \mathbb{R}$ be continuous, and therefore (Riemann-)integrable. Then $$ \forall x\in(a, b): \ \frac{d}{dx} \int_{t=a}^{x}f(t)dt=f(x). $$
Let's define $F: \ [a, b] \rightarrow \mathbb{R} $ by $F(x)=\int_{t=a}^x f(t)dt. $
Then we want to show that
$$F'(x)= \lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h}=f(x), $$
i.e. $$\forall x\in \left(a, b\right): \ \forall\varepsilon>0:\ \exists\delta>0:\ \forall t\in \underbrace{\mathcal{B}_{\delta}\left(x\right)\setminus\left\{x\right\}}_{\subset\mathbb{R}}:\ \left|\frac{F\left(x\right)-F\left(t\right)}{x-t}-f\left(x\right)\right|<\varepsilon. $$
We know that $f$ is continuous on $[a, b]$, so
$$ \forall x\in\left[a{,}\ b\right]:\ \forall\varepsilon>0:\ \exists\delta>0:\ x'\in \underbrace{\mathcal{B}_{\delta}\left(x\right)}_{\subset\left[a{,}\ b\right]}\ \Longrightarrow\ \left|f\left(x\right)-f\left(x'\right)\right|<\varepsilon.$$
Most proofs that I have seen use the mean value theorem by noting that
$$\exists c\in \underbrace{\overline{B}_{h/2}\left(x+\frac{h}{2}\right)}_{\subset\mathbb{R}}: \ \frac{F(x+h)-F(x)}{h}=f(c) \rightarrow f(x)$$as $h \rightarrow 0$. But how could this be shown rigorously by using the exact $\delta, \varepsilon$-proof?
Best Answer
Supose $h>0$ Then $\frac {F(x+h)-F(x)} h -f(x)=\frac1 h \int_x^{x+h} [f(t)-f(x)]dt$. Since $f$ is uniformly continuous, given $\epsilon >0$, there exists $\delta >0$ such that $|f(u)-f(v)| <\epsilon $ if $|u-v| \leq \delta$. Hence, $|\frac {F(x+h)-F(x)} h -f(x)| <\epsilon$ whenever $|h| <\delta$.