Chain rule for an indefinite integral after an absolutely continuous composition

If $a=b$, both conclusions are vacuous almost-everywhere assertions on a null interval. If $c=d$, then $g$ and $F\circ g$ are constant and the product is zero almost everywhere. Hence suppose $a<b$ and $c<d$.

By The indefinite integral of an $L^1$ function is absolutely continuous, $F$ is AC, and The indefinite integral of an $L^1$ function is differentiable almost everywhere gives $F^{\mathrm{\prime }}=f$ almost everywhere. Since $F$ is real-valued, Absolutely continuous functions have Luzin's property $(N)$ gives its image-null property $(N)$. The real-valued functions $g$ and $H:=F\circ g$ are AC, so Fundamental theorem of calculus for absolutely continuous functions makes both differentiable almost everywhere.

Heil's cited chain-rule theorem applies: $g$, $F$, and $H$ are differentiable almost everywhere, and $F$ maps null sets to null sets. Its conclusion holds for every function $h=F^{\mathrm{\prime }}$ almost everywhere. Taking $h=f$ yields $H^{\mathrm{\prime }}=(f\circ g)g^{\mathrm{\prime }}$ almost everywhere, including on the pullback of the exceptional set for $F^{\mathrm{\prime }}$. No assertion that this pullback is null is needed.

Since the real-valued function $H$ is AC, its derivative is integrable by Fundamental theorem of calculus for absolutely continuous functions. The product is finite-valued under our convention and equals this measurable derivative outside a null set by step 3.1. Completeness of Lebesgue measure therefore makes the product measurable, and almost-everywhere equality gives $f(g)g^{\mathrm{\prime }}\in L^1[a,b]$.