Proposition 8

Suppose g is $\mu$-integrable. If J denotes the open interval in $\mathbf{R}$ with end-points $G(a+)$ and $G(b-)$, the image under G of the measure $g \cdot \mu$ is the measure $\varphi_J \cdot \nu$ if $G(a+) \leq G(b-)$ and the measure $-\varphi_J \cdot \nu$ if $G(a+) \geq G(b-)$ (where $\nu$ denotes Lebesgue measure on G(I)).*

It suffices to prove that, for every function $f \in \mathscr{K}(G(I))$,

Now, this formula has already been proved for $g \in \mathscr{K}(I)$ (FRV, II, §2, No. 1, formula (1)). Let us pass to the general case; there exists a sequence $(g_n)$ of functions in $\mathscr{K}(I)$ such that: $1^\circ$ the sequence $(g_n(t))$ tends to $g(t)$ almost everywhere in $I$; $2^\circ$ there exists a $\mu$-integrable function $h \geq 0$ such that $|g_n| \leq h$ for all $n$ (Ch. IV, §3, No. 4, Th. 3). It follows at once from Lebesgue’s theorem that, setting $G_n(x) = c + \int_{x_0}^x g_n(t)\,dt$, the sequence $(G_n)$ converges uniformly to $G$ on $I$, and that the numbers $G_n(a+)$ and $G_n(b-)$ tend respectively to $G(a+)$ and $G(b-)$. Let $f'$ be a function in $\mathscr{K}(\mathbf{R})$ that extends $f$; the foregoing proves that $f'(G_n(t))$ tends to $f'(G(t)) = f(G(t))$ for all $t \in I$; applying Lebesgue’s theorem, one sees that the formula (12) results from the formula

by passage to the limit.