The set of total variations is not separable

Question

We know when $(X,S,\mu)$ is a measure space, $M_{\mathbb R}(S)$ consists signed (or real) measures on $(X,S)$ is Banach space. The set V consists of the measures satisfying $dv=hd\mu$ for $h \in L^{1}(\mu)$ is closed but how we can show that when $\mu$ is Lebesgue measure and S would be the cllections of Borel sets, the set V is not separable.

Answer 1

Denote $d\nu_{a}=\chi_{[0,a]}dx$ for $a>0$.

The total variation is taken as a norm, assuming that $a>b$, we compute that $\|d\nu_{a}-d\nu_{b}\|=\|\chi_{(b,a]}dx\|\geq\int\chi_{[0,1]}\chi_{(b,a]}dx=a-b$.

If we fix the $a$ and let $b<a-1$, then $\|d\nu_{a}-d\nu_{b}\|\geq 1$. But there are uncountably many such $b$, so $\{d\nu_{b}\}_{b<a-1}$ cannot be approximated by a countable subset of $M_{\mathbb{R}}$.