Lemma 6

Let $g\in \mathscr{L}^{\infty}(X, \mu)$. The spectrum of the class of $g$ in $L^{\infty}(X, \mu)$ is equal to the $\mu$-essential image of $g$.

Let us denote by $\widetilde{g}$ the class of $g$ in $L^{\infty}(X, \mu)$ and by S the $\mu$-essential image of $g$. Let $z\in \mathbf{C}-$ S. By definition, there exists an open neighbourhood U of $z$ such that $Y =\overset{-1}{g}(U)$ is locally $\mu$-negligible. The function $h$ defined by $h(x) = (g(x)-z)^{-1}$ if $x \notin Y$, and $h(x) = 0$ if $x\in Y$, then belong to $\mathscr{L}^{\infty}(X, \mu)$. Its class $\widetilde{h}$ in $L^{\infty}(X, \mu)$ satisfies $(\widetilde{g}-z)\widetilde{h}= 1$, since $(g(x)-z)h(x) = 1$ for every $x$ not belonging to the locally $\mu$-negligible set Y. Hence $z\in \mathbf{C}-$ Sp($\widetilde{g}$).

Conversely, let $z\in \mathbf{C}-$ Sp($\widetilde{g}$). Let $h\in \mathscr{L}^{\infty}(X, \mu)$ be a function whose class is the inverse of $\widetilde{g}-z$ in $L^{\infty}(X, \mu)$. There exists a real number $M>0$ such that $|h(x)|\leqslant M$ locally $\mu$-almost everywhere, and moreover one has $(g(x)-z)h(x) = 1$ locally $\mu$-almost everywhere. Let U be the open ball with center $z$ and radius $M^{-1}$ in $\mathbf{C}$; then $\overset{-1}{g}(U)$ is contained in the locally $\mu$-negligible set

therefore $z\in \mathbf{C}-$ S. The lemma is proved.