# Spectrum of $L^\infty(X,\mu)$

Suppose that $(X,\Sigma,\mu)$ is a measured set with respect to $\sigma$-algebra $\Sigma$. Suppose that $L^\infty(X,\mu)$ is the set of all $\mu$-equal bounded $\Sigma$-measurable functions on $X$. Indeed equally, one may say that $L^\infty(X,\mu)$ is the dual of $L^1(X,\mu)$. What is the spectrum of $(L^\infty(X,\mu),\|\cdot\|_\infty)$ as a Banach (C^*) algebra? Thank you very much.

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It doesn't directly affect your question, but I have a feeling that $L^\infty$ is only the dual of $L^1$ under certain mild conditions on your space and your sigma-algebra. –  Yemon Choi Jun 8 '12 at 6:45
Why the vote to close? –  Johannes Ebert Jun 8 '12 at 7:56
I've had a rough night, and might be missing something, but are you looking for the topological space which would be the Gelfand spectrum of the abelian $C*-$algebra $L^\infty$? If that's the case you should look for completely discontinuous spaces on Google, or have a look at Kadison-Ringrose, the part on VonNeumann algebras. –  Amin Jun 8 '12 at 8:26
Yes, a hyperstonean space will be the spectrum. For example, if $X=\mathbb Z$ with the counting measure, then the spectrum is $\beta Z$ (the Stone-Cech compactification). –  Yulia Kuznetsova Jun 8 '12 at 8:31
@Matthew: It's not a "standard fact"... it's a mathematical object. The OP is simply asking for extra information/intuition about that mathematical object. –  André Henriques Jun 8 '12 at 11:33

Here is a description of the spectrum of $L^\infty([0,1];\mu)$ for an arbitrary Borel measure $\mu$ on $[0,1]$.

Consider the following poset, which I call $P$ :

• The objects of $P$ are decompositions $\mathbf X=\{X_1,\ldots, X_n\}$ of $[0,1]$ into finitely many $\mu$-measurable sets $[0,1]=X_1\cup X_2\cup\ldots\cup X_n$,  $X_i\cap X_j=\emptyset$. Two decompositions $\mathbf X$ and $\mathbf Y$ are declared equal is there exists a permutation $\sigma$ such that $X_i=Y_{\sigma(i)}$ up to a $\mu$-measure zero set.
• The partial order on $P$ is given by refinement: $\mathbf X \prec \mathbf Y$ if
$Y_1=X_1\cup\ldots \cup X_{n_1}$, $Y_2=X_{n_1+1}\cup\ldots \cup X_{n_2}$, $\ldots$ (up to permutation and $\mu$-measure zero sets)

Note that the poset $P$ is filtered: given a finite set $\mathbf X_1, \mathbf X_2, \ldots, \mathbf X_n$ of elements of $P$, there is always a common refinement, i.e., an element $\mathbf X\in P$ such that $\mathbf X\prec \mathbf X_i\\,\forall i$

Given a $\mu$-measurable subset $X\subset [0,1]$, let me denote by $|X|_ \mu \subset [0,1]$ the $\mu$-adherence of $X$: $$|X|_ \mu:=\{x\in [0,1]: \forall \varepsilon>0\quad \mu(X\cap B_{x,\varepsilon})>0\},$$ where $B_{x,\varepsilon}$ denotes the ball of radius $\varepsilon$ around the point $x$. Given $\mathbf X=\{X_1,\ldots,X_n\} \in P$, we also write $|\mathbf X|_ \mu$ for the disjoint union $$|\mathbf X|_ \mu:=|X_1|_ \mu\sqcup\ldots\sqcup|X_n|_ \mu.$$

Note that if $\mathbf X \prec \mathbf Y$, then there is a natural projection map $|\mathbf X|_ \mu \twoheadrightarrow |\mathbf Y|_ \mu$.

Given the above preliminaries, the spectrum of $L^\infty([0,1];\mu)$ is given by the inverse limit of the functor $P\to Top, \mathbf X\mapsto |\mathbf X|_ \mu$:

$$Spec\big(L^\infty([0,1];\mu)\big) =\quad \underset{\mathbf X\in P}{\underset\leftarrow\lim} |\mathbf X|_ \mu$$

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