Fundamental Theorem of Calculus

Theorem. Assume that $f : [a, b] \to \Bbb{R}$ is differentiable on $[a, b]$ and $f'$ is in $L^1$. Then $$\int_{a}^{b} f'(x) \, dx = f(b) - f(a).$$

Remark. This proof is a slightly simplified version of the proof of Theorem 7.21 in Rudin's Real and Complex Analysis, 3rd edition.

Proof. Let $l$ be a lower-semicontinuous function on $[a, b]$ such that $l(x) > f'(x)$ for all $x \in [a, b]$. Define $G : [a, b] \to \Bbb{R}$ by $$G(x) = \int_{a}^{x} l(t) \, dt - [f(x) - f(a)].$$ Then for each $x \in [a, b)$, we have $$\begin{align*} \frac{G(x+h) - G(x)}{h} \geq \left( \inf_{t \in [x,x+h]}l(t) \right) - \frac{f(x+h) - f(x)}{h} \end{align*}$$ and thus $$\begin{align*} \liminf_{h\to 0^+} \frac{G(x+h) - G(x)}{h} &\geq \liminf_{h\to 0^+} l(x+h) - f'(x) \\ &\geq l(x) - f'(x) \\ &> 0. \end{align*}$$ This shows that $G$ is increasing on $[a, b)$ and by continuity, $G(b) \geq G(a) = 0$. From this, we have $$\int_{a}^{b} l(t) \, dt \geq f(b) - f(a).$$ By the Vitali-Caratheodory theorem, $f'$ can be approximated from above by lower-semicontinuous functions in $L^1$. Thus it follows that $$\int_{a}^{b} f'(t) \, dt \geq f(b) - f(a).$$ Replacing $f$ by $-f$ proves the other direction and hence the claim follows. ////