$C^\ast$-algebra theory associates a compact Hausdorff space to every measure space. What is this compact Hausdorff space? Commutative von Neumann algebras are "the same" as measure spaces, and unital commutative $C^\ast$-algebras are the same as compact Hausdorff spaces. Moreover, every commutative von Neumann algebra is in particular a unital commutative $C^\ast$ algebra. Thus every measure space 
$(X,\mu)$
has a compact Hausdorff space
$Y$
associated to it. More precisely, $Y$ is the unique compact Hausdorff space such that $C(Y) \cong L^\infty(X,\mu)$.
Question: Can one give a direct description of $Y$ in terms of $(X,\mu)$?
That is, I'd like a description which doesn't mention $C(Y)$ or $L^\infty(X,\mu)$, or any functional analysis at all, preferably.
 A: $Y$ can be described as the Stone space of the measure algebra of $(X,\mu)$.  That is, let $\Sigma$ be the $\sigma$-algebra on which $\mu$ is defined, let $N\subseteq\Sigma$ be the ideal of null sets, and let $B=\Sigma/N$ be the quotient Boolean algebra.  Then $Y$ is naturally homeomorphic to the set $S$ of Boolean homomorphisms $B\to\{0,1\}$, topologized as a subspace of $\{0,1\}^B$.
To prove this, let us first recall that $Y$ can be described as the set of $*$-homomorphisms $L^\infty(X,\mu)\to\mathbb{C}$, with the topology of pointwise convergence.  For each $b\in B$, there is a function $1_b\in L^\infty(X,\mu)$, and a $*$-homomorphism $\alpha$ must send $1_b$ to either $0$ or $1$ since $1_b^2=1_b$.  It is then easy to see that $b\mapsto \alpha(1_b)$ is a Boolean homomorphism $B\to\{0,1\}$ (the Boolean operations on sets can be expressed in terms of ring operations on their characteristic functions).  This defines a map $F:Y\to S$.
Note moreover that since simple functions are dense in $L^\infty(X,\mu)$, an element of $Y$ is determined by its values on characteristic functions $1_b$.  Thus $F$ is injective.  Also, $F$ is continuous, since the topology on $S$ is the topology of pointwise continuity with respect to evaluation at just the elements $1_b$.  Since $Y$ and $S$ are both compact Hausdorff, it follows that $F$ is an embedding.
It remains to be shown that $F$ is surjective.  Fix a homomorphism $h:B\to\{0,1\}$, and let $U=h^{-1}(\{1\})$.  The idea is that we can then define a $*$-homomorphism $L^\infty(X,\mu)\to\mathbb{C}$ which maps a function $f$ to the "limit" of the values of $f$ along the ultrafilter $U$.  To make this precise, given $f\in L^\infty(X,\mu)$ and $b\in B$, let $f[b]\subset\mathbb{C}$ denote the essential range of $f$ on $b$, and let $C_f=\{f[b]:b\in U\}$.  Note that each element of $C_f$ is compact and nonempty.  Also, $f[b\cap c]\subseteq f[b]\cap f[c]$, so $C_f$ has the finite intersection property.  Thus $\bigcap C_f$ is nonempty.  If $x\in \bigcap C_f$, then for any neighborhood $V$ of $x$ and any $b\in U$, $f^{-1}(V)\cap b$ is non-null.  Since $U$ is an ultrafilter on $B$, this means $f^{-1}(V)\in U$.  Now if we had two different points $x,y\in C_f$, they would have disjoint neighbooods $V$ and $W$, and then $f^{-1}(V)$ and $f^{-1}(W)$ would be disjoint elements of $U$.  This is impossible.
Thus, we have shown that $C_f$ has exactly one point for each $f\in L^\infty(X,\mu)$.  Define $\alpha(f)$ to be the unique element of $C_f$, which can also be described as the unique point $x$ such that given any neighborhood $V$ of $x$, for all sufficiently small $b\in U$, $f|_b$ takes values in $V$ almost everywhere.  This description makes it easy to verify that $\alpha$ is a $*$-homomorphism, and that $\alpha(1_b)=h(b)$ for each $b\in B$.  Thus $\alpha\in Y$ and $h=F(\alpha)$, so $h$ is in the image of $F$, as desired.
(Alternatively, to show $F$ is surjective, by Stone duality it suffices to show that the image of $F$ separates elements of $B$, since closed subspaces of the Stone space $S$ correspond to quotients of the algebra $B$.  But by Gelfand duality, elements of $Y$ separate elements of $L^\infty(X,\mu)$, and so we're done since distinct elements of $B$ have distinct characteristic functions in $L^\infty(X,\mu)$.)
