When is $(L^\infty (\mathbb{R}, \mu); \|\cdot\|_{L^\infty (\mathbb{R}, \mu)})$ separable?

Question

Let $\mu$ be a measure on the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$. What would be a necessary and sufficient condition on $\mu$ s.t. $(L^\infty (\mathbb{R}, \mu); \|\cdot\|_{L^\infty (\mathbb{R}, \mu)})$ is separable?

Answer 1

If $\mu = \sum\limits_{k=1}^{n} a_k\delta_{x_k}$ for some $n, a_k,x_k$'s then $L^{\infty} (\mu)$ is finite dimensional, Hence separable. Let us show that this condition is also necessary. First observe the following: if there exists a disjoint sequence $(A_n)$of Borel sets with $\mu(A_n)>0$ for all $n$ then $L^{\infty} (\mu)$ is not separable. This is because $\|I_A-I_B\|_{L^{\infty}} =1$ whenever $A$ and $B$ are unions of $A_n$'s over distinct index sets, so we would have uncountable many functions separated by distance $1$. Now looking at the intervals $[n,n+1),n\in \mathbb Z$ conclude that $\mu$ is supported by some compact interval $[a,b]$. Next note that for any real number $a$ the intervals $(a-\frac 1 n, a-\frac 1 {n+1})$ have measure $0$ for $n$ sufficiently large and a similar result holds for the intervals $(a+\frac 1 {n+1}, a+\frac 1 n)$. Hence, for each $a$ there exists $\delta >0$ such that $\mu ((a-\delta, a+\delta) \setminus \{a\})=0$. I leave the rest of the argument to you because you only need to use compactness of $[a,b]$ to complete the proof.