Lemma 2

Let X be a locally compact topological space and let $\mu$ be a positive measure on X.

a) Let $g$ be a $\mu$-measurable function on X. The multiplication operator $m_g$ on $L^2(X, \mu)$ is normal and $b(m_g) =m_{\beta\circ g}$;

b) Let $h: X\rightarrow D$ be a $\mu$-measurable function. The endomorphism $m_h$ belongs to $\Omega (L^2(X, \mu))$ and $B(m_h) =m_{\beta^{-1}\circ h}$.

The partial operator $m_g$ is closed with dense domain. Since $m^*_gm_g=$ $m_{\overline{g}}m_g=m_{|g|^2}$ (prop. 23 of IV, p. 253 and prop. 24 of IV, p. 255), one has

(prop. 22 of IV, p. 252). This implies

by formula (2) of IV, p. 262, cor. of prop. 6 of IV, p. 187 and lemma 6 of IV, p. 254. Finally applying prop. 6 of IV, p. 187, one obtains assertion a).

Let us prove b). Since $1- |h(x)|^2>0$ for every $x\in X$, one has $m_h\in$ $\Omega (L^2(X, \mu))$ (cor. of Prop. 6 of IV, p. 187). Since the mapping $u\mapsto b(u)$ is injective (Prop. 2 of IV, p. 264) and the endomorphisms $b(B(m_h))$ and $b(m_{\beta^{-1}\circ h})$ are equal to $m_h$ by assertion a), it follows that $B(m_h) =m_{\beta^{-1}\circ h}$.