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Housekeeping: moved von Neumann algebras to its own part.

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Bokuan Li
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\section{$L^\infty$}
\label{section:l-infty-algebra}
\begin{proposition}
\label{proposition:measures-dual-algebra}
Let $X$ be a compact Hausdorff space and $\mathscr{M} \subset M_R(X; \complex)$ be a closed subspace such that:
\begin{enumerate}
\item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
\end{enumerate}
and
\[
J: C(X; \complex) \to \mathscr{M}^* \quad \dpn{\mu, J(f)}{\mathscr{M}} = \int f d\mu
\]
then
\begin{enumerate}
\item $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$.
\item There exists a unique weak*-continuous involution on $\mathscr{M}^*$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by
\[
\dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
\]
\item There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
\item $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^*$-algebra.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using \autoref{theorem:hilbert-measures-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, identify
\[
\mathscr{M} = [l^1(I); L^1(\mu_i; \complex)] \quad \mathscr{M}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]
\]
(1): Under the above, $C(X; \complex)$ may be identified as the diagonal
\[
\bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I} \subset [l^\infty(I); L^\infty(\mu_i; \complex)]
\]
By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. As a result, $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$.
(2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$,
\[
\dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
\]
so the conjugation map is weak*-continuous.
(3): Let $f, g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$ and $\mu \in [l^1(I); L^1(\mu_i; \complex)]$,
\[
\dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]}
\]
so the composition map is separately weak*-continuous.
(4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra.
\end{proof}
\begin{lemma}
\label{lemma:linfty-multiply-bound}
Let $(X, \cm, \mu)$ be a measure space, $f: X \to \complex$ be a measurable function, and $C \ge 0$ such that $\norm{fg}_{L^2(X; \complex)} \le C\norm{g}_{L^2(X; \complex)}$ for all $g \in L^2(X; \complex) \cap \Sigma(X; \complex)$, then $f \in L^\infty(X; \complex)$ with $\norm{f}_{L^\infty(X; \complex)} \le C$.
\end{lemma}
\begin{proof}
For any $h \in L^1(X; \complex) \cap L^\infty(X; \complex)$,
\begin{align*}
\normn{f^2h}_{L^1(X; \complex)} &= \int_X |f^2| \cdot |h| = \int_X |f||h|^{1/2} \cdot |f||h|^{1/2} = \normn{f|h|^{1/2}}_{L^2(X; \complex)}^2 \\
&\le C^2 \normn{|h|^{1/2}}_{L^2(X; \complex)}^2 = C^2 \norm{h}_{L^1(X; \complex)}
\end{align*}
By \autoref{theorem:lp-dual-function}, $f^2 \in L^\infty(X; \complex)$ with $\normn{f^2}_{L^\infty(X; \complex)} \le C^2$. Hence $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le C$.
\end{proof}
\begin{proposition}
\label{proposition:linfty-von-neumann-algebra}
Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
\begin{enumerate}
\item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
\item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
\item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is maximal abelian.
\end{enumerate}
\end{proposition}
\begin{proof}
(2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$.
(3): Using \autoref{proposition:maximal-commutative-vn}, it is sufficient to show that $L^\infty(X; \complex)' = L^\infty(X; \complex)$. For each $g \in L^\infty(X; \complex)$, denote $M_g \in B(L^2(X; \complex))$ as the multiplication map by $g$.
First suppose that $\mu(X) < \infty$ and let $f = T\one_X$, then for each $g \in L^\infty(X; \complex)$,
\[
M_fg = M_g f= M_g T\one_X = TM_g\one_X = Tg
\]
By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex) \subset L^\infty(X; \complex)$ is dense in $L^2(X; \complex)$, so $M_fg = Tg$ for all $g \in L^2(X; \complex)$.
Now suppose that $\mu$ is an arbitrary localisable measure. For each $A \in \cm$ with $\mu(A) < \infty$, the finite case shows that there exists $f_A \in L^\infty(X; \complex)$ such that
\begin{enumerate}[label=(\roman*)]
\item $f_A|_{X \setminus A} = 0$ almost everywhere.
\item $Tg = M_{f_A}g$ for all $g \in L^2(X; \complex)$ with $f_A|_{X \setminus A} = 0$ almost everywhere.
\end{enumerate}
By (ii), $f_A|_{A \cap B} = f_B|_{A \cap B}$ for all $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$. By the \hyperref[gluing lemma for meaasurable functions]{lemma:gluing-measurable}, there exists $f \in L^\infty(X; \complex)$ such that $f|_A = f_A|_A$ for all $A \in \cm$ with $\mu(A) < \infty$.
For each $g \in L^2(X; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. As such, there exists $\seq{A_n} \subset \cm$ such that $\bracsn{g \ne 0} = \bigsqcup_{n \in \natp}A_n$ and $\mu(A_n) < \infty$ for all $n \in \natp$. In which case,
\[
Tg = \sum_{n = 1}^\infty T(\one_{A_n}g) = \sum_{n = 1}^\infty M_{f_{A_n}} \one_{A_n}g = M_fg
\]
By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. Therefore $T = M_f$, and $L^\infty(X; \complex)$ is maximal abelian.
\end{proof}
\begin{theorem}[Order Structure of $L^\infty$]
\label{theorem:l-infty-dedekind-complete}
Let $(X, \cm, \mu)$ be a localisable measure space, then
\begin{enumerate}
\item $L^\infty(X; \real)$ is order complete.
\item For each $\phi \in L^1(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^\infty(X; \real)$,
\[
\sup_{f \in S} \dpn{\phi, f}{L^1(X; \real)} = \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}
\]
\end{enumerate}
\end{theorem}
\begin{proof}
By \autoref{proposition:linfty-von-neumann-algebra}, $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
(1): Since $L^\infty(X; \real)$ is a lattice, it is order complete by \autoref{theorem:existence-of-projections-vna}.
(2): By \autoref{theorem:existence-of-projections-vna}, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of \autoref{proposition:linfty-von-neumann-algebra}, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)} = \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.
\end{proof}
\begin{corollary}
\label{corollary:linfinity-extremally-disconnected}
Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremally disconnected.
\end{corollary}
\begin{proof}
By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremally disconnected.
\end{proof}
\begin{lemma}
\label{lemma:l-infty-order-isotone}
Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^\infty(X; \complex)\to L^\infty(Y; \complex)$ such that:
\begin{enumerate}[label=(\alph*)]
\item $T$ is an isometric isomorphism.
\item For each $f, g \in L^\infty(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$.
\end{enumerate}
then:
\begin{enumerate}
\item $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$.
\item $T$ is weak*-continuous.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\phi \in L^1(Y; \complex)$ with $\phi \ge 0$, then $T^*\phi \in L^\infty(X; \complex)^*$. For each $\seq{B_n} \subset \cm$ and $B \in \cm$ with $B_n \upto B$, $\one_B = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^\infty(X; \complex)$. Thus (b) and \autoref{theorem:l-infty-dedekind-complete} imply that
\begin{align*}
\sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)} &= \sup_{n \in \natp} \dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)} = \dpn{\phi, T\one_B}{L^1(Y; \complex)} \\
&= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}
\end{align*}
Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(X; \complex)$ with $f \ge 0$ such that $\int_B f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$.
Let $g \in L^\infty(X; [0, 1])$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists simple functions $\seq{g_n} \subset \Sigma(X; [0, 1])$ such that $g_n \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_n$ as an element of $L^\infty(X; \real)$, so (b) and \autoref{theorem:l-infty-dedekind-complete} imply that,
\begin{align*}
\dpn{\phi, Tg}{L^1(Y; \complex)} &= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)} = \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)} \\
&= \dpn{f, g}{L^1(X; \complex)}
\end{align*}
By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)} = \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^\infty(X; \complex)$. Therefore $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$, and $T$ is weak*-continuous.
\end{proof}
\begin{theorem}[Uniqueness of $L^\infty$]
\label{theorem:linfty-uniqueness}
Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_X \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^\infty(\mu; \complex) \to L^\infty(\nu; \complex)$ such that:
\begin{enumerate}[label=(\alph*)]
\item $\Phi$ is a *-isomorphism.
\item $\Phi|_{C(X; \complex)}$ is the identity.
\end{enumerate}
then:
\begin{enumerate}
\item $\mu$ and $\nu$ are equivalent.
\item $L^\infty(\mu; \complex) = L^\infty(\nu; \complex)$.
\item $\Phi$ is the identity map.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 21.4]{Zhu}}}. ]
As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^\infty(\mu; \complex)$.
(1): By \autoref{lemma:l-infty-order-isotone}, there exists $f \in L^1(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0} implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent.
(3): By \autoref{lemma:l-infty-order-isotone}, $\Phi$ is also weak*-continuous. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $L^\infty(\mu; \complex)$ and $L^\infty(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^\infty(\mu; \complex)$.
\end{proof}