Change of variables for an increasing absolutely continuous function

First take $f={\mathbf{1}}_{(r,s)}$. The clipped function $(g-r{)}^{+}-(g-s{)}^{+}$ is AC; its derivative is ${\mathbf{1}}_{(r,s)}(g)g^{\mathrm{\prime }}$ almost everywhere (on a level set of an AC increasing function, $g^{\mathrm{\prime }}=0$ almost everywhere). The sharp FTC evaluates its integral as the length of $(r,s)\cap [g(a),g(b)]$.

The interval-indicator identity in step 1.1 extends first to the algebra of finite unions of intervals and then, by The monotone class generated by an algebra equals the sigma-algebra it generates, to all Borel indicators. If $N$ is Lebesgue null, Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it covers it by open sets $O$ of arbitrarily small length. The open-set case then makes $g^{-1}(N)\cap \{g^{\mathrm{\prime }}>0\}$ null: on $\{g^{\mathrm{\prime }}\ge 1\mathrm{/}k\}$ its outer measure is at most $k{\int }_{g^{-1}(O)}{g}^{\mathrm{\prime }}\le k\lambda (O)$, and take the union over $k$. Every Lebesgue-measurable $f$ has a Borel representative off a null set, so this observation makes $f(g)g^{\mathrm{\prime }}$ agree almost everywhere with a measurable weighted composition. Simple-function approximation and Monotone convergence for the integral now extend the identity to every nonnegative measurable $f$.

Apply step 2.1 to $f^{+}$ and $f^{-}$. Applying it also to $\mathrm{\mid }f\mathrm{\mid }$ gives ${\int }_a^b\mathrm{\mid }f(g(x))\mathrm{\mid }{g}^{\mathrm{\prime }}(x)dx={\int }_{g(a)}^{g(b)}\mathrm{\mid }f(y)\mathrm{\mid }dy<\mathrm{\infty }$, so $f(g)g^{\mathrm{\prime }}\in L^1$ and subtraction gives the displayed formula. Constant and singleton cases have both integrals zero. This is also the conclusion of the cited Heil corollary.