Lemma 1

Let $g$ be a $\mu$-measurable function from X into a topological space Y, and S its $\mu$-essential image. The set of elements $x\in X$ such that $g(x)$ does not belong to S is locally $\mu$-negligible in X.

Let $Z =\overset{-1}{g}(Y$ - S) be the set in question.

First suppose that X is compact and $g$ continuous. In this case, the image measure $g(\mu)$ is defined (INT, V, p. 69, § 6, n$^o1$, def. 1) since $\mu$ is a bounded measure. It follows from the definitions that the $\mu$-essential image of $g$ is the support of the measure $g(\mu) ($cf. INT, V, p. 70, § 6, n$^o2$, cor. 2), whence $\mu(Z) =g(\mu)(Y$ - S) = 0 (INT, IV, p. 118, § 2, n$^o2$, prop. 5).

Now consider the general case. Let C be a compact subset of X, and let $\varepsilon >0$ be a real number. Since $g$ is measurable, there exists a compact subset $C_1\subset C$ such that $\mu(C$- $C_1)\leqslant \varepsilon$ and such that $g|C_1$ is continuous (INT, IV, p. 169, § 5, n$^o1$, prop. 1). One then has

$\mu(Z\cap C)\leqslant \mu(C$ - $C_1) +\mu(Z\cap C_1)\leqslant \varepsilon +\mu(Z\cap C_1)$.

Let $\mu_1$ be the measure induced by $\mu$ on $C_1$ (INT, IV, p. 186, § 5, n$^o7$, def. 4). Let $S_1$ be the $\mu_1$-essential image of $g|C_1$. One has $Z\cap C_1\subset$ $(g|C_1)^{-1}(Y$ - $S_1)$. By the first case, the set $Z\cap C_1$ is therefore $\mu_1$-negligible, and consequently $\mu$-negligible (INT, IV, p. 187, § 5, n$^o7$, lemma 2, (i)). The above inequality becomes $\mu(Z\cap C)\leqslant \varepsilon$; since $\varepsilon$ and C are arbitrary, the set Z is locally $\mu$-negligible (INT, IV, p. 172, § 5, n$^o2$, prop. 5).