Total-variation function of an absolutely continuous function

The sharp FTC gives $F(v)-F(u)={\int }_u^v{F}^{\mathrm{\prime }}$ for $a\le u\le v\le b$. Hence every partition sum on $[u,v]$ is at most ${\int }_u^v\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$, so the variation on every subinterval is finite.

Variation is additive across an inserted endpoint: refining any partition of $[a,v]$ at $u$ gives one inequality, while joining partitions of $[a,u]$ and $[u,v]$ gives the reverse inequality after taking suprema. Thus, writing $H(x)={\int }_a^x\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$, $0\le V_F(v)-V_F(u)=V(F,[u,v])\le H(v)-H(u)\mathrm{.}$ The indefinite integral $H$ is absolutely continuous by The indefinite integral of an $L^1$ function is absolutely continuous. For every finite disjoint family of short intervals, the displayed inequality bounds the sum of their $V_F$-increments by the corresponding sum of $H$-increments, so $V_F$ is absolutely continuous as well.

By The indefinite integral of an $L^1$ function is differentiable almost everywhere, $H^{\mathrm{\prime }}=\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ almost everywhere. At every common differentiability point of $F$, $V_F$ and $H$, the definition of variation and step 2.1 give, for $h>0$, $\mathrm{\mid }F(x+h)-F(x)\mathrm{\mid }\le V_F(x+h)-V_F(x)\le H(x+h)-H(x)\mathrm{.}$ Dividing by $h$ and letting $h\downarrow 0$ yields $\mathrm{\mid }{F}^{\mathrm{\prime }}(x)\mathrm{\mid }\le V_F^{\mathrm{\prime }}(x)\le H^{\mathrm{\prime }}(x)=\mathrm{\mid }{F}^{\mathrm{\prime }}(x)\mathrm{\mid }$. Thus $V_F^{\mathrm{\prime }}=\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ almost everywhere. Since $V_F$ is absolutely continuous and $V_F(a)=0$, the sharp FTC Fundamental theorem of calculus for absolutely continuous functions gives $V_F(x)={\int }_a^x{V}_F^{\mathrm{\prime }}(t)dt={\int }_a^x\mathrm{\mid }{F}^{\mathrm{\prime }}(t)\mathrm{\mid }dt$ for every $x$.