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.



Spectrum of $L^\infty(X,\mu)$
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 $P\to Top, \mathbf X\mapsto |\mathbf X|\_ \mu$:



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


