Lemma 5

Let $\mu$ be a positive measure on X. Let $g$ be a $\mu$-measurable function from X into $\mathbf{C}$, and let S be its $\mu$-essential image. For every function $f\in \mathscr{L}_u(S)$, the mapping $h$ from X into $\mathbf{C}$ such that $h(x) = 0$ if $g(x)\notin S$ and $h(x) =f(g(x))$ if $g(x)\in S$ is $\mu$-measurable.

Let $x\in X$ and let U be a relatively compact open neighbourhood of $x$. The set N = U - $(U\cap \overset{-1}{g}(S))$ is $(\mu|$U)-negligible. For every integer $n\geqslant 1$, let $V_n$ be an open set such that $N\subset V_n\subset U$ and $\mu(V_n-N)<1/n$ (INT, IV, p. 116, § 1, n$^o4$, prop. 19).

The restriction $\widetilde{g}$ of $g$ to U- $V_n$ is $\mu|(U$- $V_n$)-proper, and its image is contained in S. Since $f$ is universally measurable, the mapping $x\mapsto f(g(x))$ from U - $V_n$ into $\mathbf{C}$ is measurable with respect to the measure $\mu|(U$ - $V_n)$ (INT, V, p. 71, § 6, n$^o2$, prop. 3). The mapping $h_n$ from X into $\mathbf{C}$ such that $h_n(x) = 0$ if $x \notin U$ - $V_n$ and $h_n(x) =f(g(x))$ if $x\in U$ - $V_n$ is therefore $\mu$-measurable (INT, IV, p. 193, § 5, n$^o10$, prop. 16).

Since $h_n(x)\rightarrow h(x)$ for $\mu$-almost all $x\in U$, the restriction of $h$ to U is $\mu$-measurable (INT, IV, p. 175, § 5, n$^o4$, th. 2). It follows that $h$ is $\mu$-measurable by the localisation principle (INT, IV, p. 171, § 4, n$^o2$, prop. 4).