Corollary

If a function $\mathbf{f}$ defined on $G(I)$, with values in $\overline{\mathbf{R}}$ or in a Banach space, is such that the function $t \mapsto \mathbf{f}(G(t))g(t)$ is integrable on $I$ for Lebesgue measure, then $\mathbf{f}$ is integrable on $J$ for Lebesgue measure and

(formula for change of variable in the Lebesgue integral).

For, $\mathbf{f}(G(t))$ is integrable for the measure $|g|\cdot\mu$, hence also for the measures $g^+\cdot\mu$ and $g^-\cdot\mu$; it follows from Th. 1 (No. 2) that $\mathbf{f}$ is integrable for the image measures $G(g^+\cdot\mu)$ and $G(g^-\cdot\mu)$, hence also for the measure $\varphi_J \cdot \nu$, and that (13) holds, on taking into account Prop. 8 and formula (9).

It can happen that $\mathbf{f}$ is integrable on $J$ for Lebesgue measure, but that $t \mapsto \mathbf{f}(G(t))g(t)$ is not integrable on $I$ for Lebesgue measure (Exer. 10).

Now suppose that $g$ maintains a constant sign almost everywhere (and is locally $\mu$-integrable); one may suppose for example that $g(t) \geq 0$ almost everywhere in $I$. Then $G$ is an increasing continuous function on $I$, therefore $G(a+)$ and $G(b-)$ exist (but may be infinite). Moreover, $G$ is a $\lambda$-proper mapping of $I$ into $G(I)$: for, if $G(b-) \in G(I)$, there is an $x_1 \geq x_0$ such that $G$ is constant for $x \geq x_1$, and then the inverse image under $G$ of the compact interval $[G(x_0), G(b-)]$ is $\lambda$-integrable; if, on the contrary, $G(b-) \notin G(I)$, then the inverse image under $G$ of every compact interval with left end-point $G(x_0)$, contained in $G(I)$, differs from a compact interval by at most a $\lambda$-negligible interval. One argues similarly for the compact intervals with right end-point $G(x_0)$, whence our assertion. Moreover: