The indefinite integral of an $L^1$ function is differentiable almost everywhere

Extend $f$ by $0$ outside $[a,b]$, obtaining a function $g$ on [L1, given, construct, algebra] $\mathbb{R}$. Because $g$ is bounded by $\mathrm{\mid }f\mathrm{\mid }$ on $[a,b]$ and vanishes elsewhere, ${\int }_{\mathbb{R}}\mathrm{\mid }g(t)\mathrm{\mid }dt={\int }_a^b\mathrm{\mid }f(t)\mathrm{\mid }dt<\mathrm{\infty },$ so $g\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$. For $x\in (a,b)$ and every $r>0$, define $E_r^{+}(x):=[x,x+r),E_r^{-}(x):=(x-r,x]\mathrm{.}$ Each family shrinks nicely to $x$ with constant $\alpha =\frac{1}{2}$, because $E_r^{\pm }(x)\subseteq B(x,r)$ and $\lambda (E_r^{\pm }(x))=r=\frac{1}{2}\lambda (B(x,r))$.

Apply [L2] to the set $A:=(a,b)$ and the family $E_r^{+}(x)$ from step [L2, step 1.1, algebra] 1.1. This gives a full-measure subset $A_{+}\subseteq (a,b)$ such that ${\lim }_{r\to 0^{+}}\frac{1}{r}{\int }_x^{x+r}g(t)dt=g(x)(x\in A_{+})\mathrm{.}$ Applying [L2] again to the family $E_r^{-}(x)$ gives another full-measure subset $A_{-}\subseteq (a,b)$ such that ${\lim }_{r\to 0^{+}}\frac{1}{r}{\int }_{x-r}^{x}g(t)dt=g(x)(x\in A_{-})\mathrm{.}$ Hence both one-sided limits hold for every $x\in A_{+}\cap A_{-}$, which still has full measure in $(a,b)$. At such an $x$, if $h>0$ is small then $\frac{F(x+h)-F(x)}{h}=\frac{1}{h}{\int }_x^{x+h}f(t)dt=\frac{1}{h}{\int }_x^{x+h}g(t)dt,$ while for $h<0$, $\frac{F(x+h)-F(x)}{h}=\frac{1}{\mathrm{\mid }h\mathrm{\mid }}{\int }_{x+h}^{x}g(t)dt\mathrm{.}$ Both one-sided limits therefore equal $g(x)=f(x)$.

Take $E=A_{+}\cap A_{-}$ from step 2.1. It is measurable and has full measure in $(a,b)$, and $F^{\mathrm{\prime }}(x)=f(x)$ for every $x\in E$. If $\tilde{f}=f$ almost everywhere, then their integrals on every interval $[a,x]$ agree, so they define the same $F$. Removing also the null set on which their values differ gives $F^{\mathrm{\prime }}=\tilde{f}$ on a full-measure set. Thus the conclusion holds for every fixed representative.