When is $L^\infty(E)$ separable? I wonder how to exhibit a measurable set $E \subset \mathbb{R}$ for which $L^\infty(E)$ is separable. It's clear that $L^\infty(E)$ is not separable if $E$ contains any nondegenerate interval(to be honest i am still a little confused about the result ). Any idea? 
 A: If the measure you're using has to  be Lebesgue measure, then you're out of luck. The only measurable sets $E$ for which $L^\infty(E)$ is separable are the sets of measure zero, for which $L^\infty(E)$ is the zero vector space.
If you allow other measures, then you could let $E$ be a finite set, with some measure that gives each point in $E$ a positive measure.  Then $L^\infty(E)$ is finite-dimensional and thus separable.  But this is essentially all you can do.  
Here's a proof for the case of Lebesgue measure; it can be adapted to other situations. If $E$ has positive measure, it can be partitioned into a countable infinity of measurable subsets $A_0,A_1,\dots$ that all have positive measure.  To each subset $X$ of $\mathbb N$, associate the function $f_X$ that sends all points from $A_n$ to $1$ if $n\in X$ and  to $0$ otherwise.  Each of these functions $f_X$ is in $L^\infty(E)$, and the $L^\infty$ distance between any two of them is $1$. So we have uncountably many elements of $L^\infty(E)$, all a distance $1$ from each  other.  That prevents separability, since no point can be closer than a distance $\frac12$to more than one of these points $f_X$.
