For $\delta > 0$: \begin{align*} \left|\frac{F(x+\delta) - F(x)}{\delta} - f(x)\right| = \frac{1}{|\delta|} \left|\int_x^{x+\delta} (f(y) - f(x)) \, d\mathcal{L}^1(y)\right| \le \frac{1}{|\delta|} \int_x^{x+\delta} |f(y) - f(x)| \, d\mathcal{L}^1(y). \end{align*} Since $B_\delta(x) = (x - \delta, x + \delta)$ has $\mathcal{L}^1(B_\delta(x)) = 2\delta$, the right-hand side is bounded by \begin{align*} \frac{2}{\mathcal{L}^1(B_\delta(x))} \int_{B_\delta(x)} |f(y) - f(x)| \, d\mathcal{L}^1(y) \to 0 \quad \text{as } \delta \to 0 \end{align*} for a.e. $x$ by the [Lebesgue Differentiation Theorem](/theorems/74). The same argument applies for $\delta < 0$. Hence $F'(x) = f(x)$ for a.e. $x$.