Proof, [Theorem 1.47, Fol16]. (1): Fix $i \in I$ and let $\mu_{i} = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_{i}(C) = 0$. By (1) and (2) of Lemma 37.4.2, $\mu_{i}$ is a finite positive Radon measure.

By Spectral Theorem I, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$,

so $\Gamma_{A}S \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i}= dE_{S\xi_i, T\xi_i}\ll \mu_{i}$. By (1) of Lemma 37.4.2 and completeness of $L^{1}(\mu; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_{i}$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_{i}(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$.

(2): Let

For any $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$,

because $\ol{A\xi_i}\perp \ol{A\xi_j}$, so $E_{x, y}= 0$. By (1), $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}\subset S([l^{1}(I); L^{1}(\mu_{i}; \complex)])$ for all $i \in I$. For each $i \in I$, let $P_{i} \in B(H)$ be the orthogonal projection of $H$ onto $\ol{A\xi_i}$, then as $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_{i}x$ for all $x \in H$. Thus for any $x, y \in H$,

so $\mathscr{E}\subset S([l^{1}(I); L^{1}(\mu_{i}; \complex)])$.

On the other hand, for each $i \in I$, since $\mu_{i}$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^{1}(\mu_{i}; \complex)$ by Proposition 25.1.7. As

and $\mathscr{E}\subset M_{R}(\Omega(A); \complex)$ is closed, $\mathscr{E}\supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$.

Finally, given that the above holds for all $i \in I$, $\mathscr{E}= S([l^{1}(I); L^{1}(\mu_{i}; \complex)])$. By Theorem 16.5.6 and Theorem 16.3.4, $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)] = [l^{1}(I); L^{1}(\mu_{i}; \complex)]^{*}$, so $\mathscr{E}^{*}$ may be identified with its image under the adjoint of $S$.

(3): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_{i} = T\xi_{i}$,

so $\Gamma_{A} S = \Gamma_{A} T$ $\mu_{i}$-almost everywhere. Thus the mapping

is well-defined. Moreover, for any $S, T \in A$,

so $U_{i}$ extends into an isometry between $\ol{A\xi_i}$ and $L^{2}(\mu_{i}; \complex)$.

For each $i \in I$, let $P_{i} \in B(H)$ be the orthogonal projection of $H$ onto $\ol{A\xi_i}$, then

is an isometry between $H$ and $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$ such that $U(Tx) = \Gamma_{A}T \cdot Ux$ for all $x \in H$ and $T \in A$.

Finally, given that

1. (i)

   By Goldstine’s Theorem, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^{*}$.

2. (ii)

   The weak* topology on $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ is equal to the weak operator topology of $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ acting on $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$.

3. (iii)

   $A$ is weak-operator dense in $B$.

4. (iv)

   By Spectral Theorem I, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B$.

the mapping $U$ is a unitary equivalence between $\mathscr{E}^{*}$ acting on $[l^{2}(I); L^{2}(\mu_{i}; \complex)]$ and $B$ acting on $H$.$\square$