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Let $(X, \mathcal A, \mu)$ be a measure space. Let $\mathcal F$ be the collection of all measurable sets with finite measure. We define a pseudometric metric $d_\mu$ on $\mathcal F$ by $d_{\mu}(A, B) := \mu(A \triangle B)$. Then $d_\mu$ becomes a metric when $\mathcal F$ is considered modulo the equivalence relation $\sim$ defined by $$ A \sim B \iff \mu(A \triangle B) = 0 \quad \forall A,B \in \mathcal F. $$

In a previous thread, I proved that

Theorem: If $\mathcal A$ is countably generated and $\mu$ is $\sigma$-finite, then $d_\mu$ is separable.

I would like to ask if the result still holds without $\sigma$-finiteness assumption, i.e.,

If $\mathcal A$ is countably generated, then $d_\mu$ is separable.

Akira
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No. Take $\mathbb{R}$ with the Borel sets and let $\mu$ be the counting measure, so that $\mu(B)$ is $\# B$ for $B$ finite and $\mu(B)=\infty$ otherwise. For $x,y\in\mathbb{R}$ with $x\neq y$, we have $d_\mu\big(\{x\},\{y\}\big)=2$. So you have an uncountable set of elements of distance $2$. This implies that there is no countable dense subset.

Michael Greinecker
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