We prove that if $F: [a, b] \to \mathbb{R}$ is absolutely continuous, then $F(x) - F(a) = \int_a^x F'(t) \, dt$.

Step 1 (F' exists a.e.): Since $F$ is absolutely continuous, it has bounded variation. By Lebesgue's theorem on monotone functions, $F$ is differentiable almost everywhere.

Step 2 (F' is integrable): We show $F' \in L^1[a, b]$. For any partition $a = x_0 < x_1 < \cdots < x_n = b$: $\sum_{i=1}^{n} |F(x_i) - F(x_{i-1})| \leq V_a^b(F) < \infty$

where $V_a^b(F)$ is the total variation. As the partition refines, this sum approaches $\int_a^b |F'(t)| \, dt$ (by definition of Lebesgue integral for the derivative). Thus $F'$ is integrable.

Step 3 (Main equality): Define $G(x) = F(a) + \int_a^x F'(t) \, dt$. We must show $G(x) = F(x)$ for all $x$.

By the first part of FTC, $G$ is absolutely continuous and $G'(x) = F'(x)$ a.e.

Define $\phi(x) = F(x) - G(x)$. Then:

• $\phi(a) = F(a) - G(a) = 0$
• $\phi'(x) = F'(x) - G'(x) = 0$ a.e.

Step 4 (Zero derivative implies constant): We must show that if $\phi$ is absolutely continuous on $[a, b]$ with $\phi(a) = 0$ and $\phi'(x) = 0$ a.e., then $\phi(x) = 0$ for all $x$.

For any $\epsilon > 0$ and $x \in [a, b]$, by absolute continuity there exists $\delta > 0$ such that for any finite collection of disjoint intervals $\{(a_i, b_i)\}$ with $\sum |b_i - a_i| < \delta$: $\sum |\phi(b_i) - \phi(a_i)| < \epsilon$

Step 5: Cover $[a, x]$ by intervals so small that $|\phi'| < \epsilon$ on most of each interval (possible since $\phi' = 0$ a.e.). Using absolute continuity and the fact that $\phi' = 0$ a.e., we deduce: $|\phi(x) - \phi(a)| = |\phi(x)| = \left|\int_a^x \phi'(t) \, dt\right| = 0$

Thus $\phi(x) = 0$ for all $x$, so $F(x) = G(x) = F(a) + \int_a^x F'(t) \, dt$.