Connection between separable measure spaces and $\sigma$-finite measure spaces

Question

I recently came across a theorem which makes a hypothesis that a certain measure space is separable (the definition can be found here). In order to avoid confusion, I'll add the definition here:

We have a space $X$ with measure. Then we define a distance on measurable sets $A$ and $B$ as $d(A,B)=meas(A\triangle B) = meas ((A\setminus B)\cup(B\setminus A))$. The space is called separable measure space iff the space of measurable subsets is separable with respect to distance $d$. In other words, there exists a countable sequence $S$ of measurable subsets of $X$, such that for all measurable $A\subset X$ and $\forall \delta>0$ there exists $B\in S$ such that $meas(A\triangle B)<\delta $.

On the other hand, it's established that in non-$\sigma$-finite spaces some weird things can happen (Fubini's theorem no longer valid, etc. This wiki article briefly touches this subject). Therefore, I'd like to have $\sigma$-finiteness.

We know that $L^p$ spaces for finite $p$ over separable measure spaces are separable; on the other hand, we can build a nonseparable $L^2$ space over a $\sigma$-finite measure space (see this answer), therefore $\sigma$-finiteness does not imply separability (in the sense of measures).

My intuition suggests that separability of the measure space implies $\sigma$-finiteness, however, I'm not able to prove or disprove it (this doesn't immediately follow from the definition).

I'd be grateful if you could provide me with references where such things are discussed. And hints on how to prove/disprove the hypothesis above are welcome, too.

Edit

One can establish that a semifinite separable measure is $\sigma$-finite, as was indicated in now deleted answer by Norbert. Moreover, the accepted answer by Michael Greinecker shows that a non-semifinite separable measure space can be non-$\sigma$-finite.

Answer 1

Separable measure spaces do not have to be $\sigma$-finite. Let $X=\{0\}$, $\Sigma=\{X,\emptyset\}$, and let $\mu(X)=\infty$. Then $\Sigma$ is also a countable dense subset.

Answer 2

Since I was not able to digest the result by Fremlin in Norbert's answer, I will give a simple proof.

Let us assume that $\mu$ is separable and semifinite. Let $\mathcal{E} \subset \Sigma$ be the countable set given by the separability of $\mu$. We define $$ \mathcal E_0 := \{ E \in \mathcal E \mid \mu(E) < \infty\} $$ and $$ E_0 := \bigcup_{E \in \mathcal E_0} E.$$ It remains to show that $E_1 := X \setminus E_0$ is a $\mu$-null set. Towards a contradiction, suppose that $\mu(E_1) > 0$. By semifiniteness, there exists a measurable $E_2 \subset E_1$ with $\mu(E_2) \in (0,\infty)$. Thus, there exists $E_3 \in \mathcal E$ with $\mu( E_2 \mathbin{\Delta} E_3) < \mu(E_2)/2$. Thus, $E_3 \in \mathcal E_0$ and this is a contradiction.

This shows that $\mu$ is $\sigma$-finite.

Answer 3

Answer undeleted by request

I think this proposition 215B (iv) in Measure Theory. Vol 2. Broad foundations. D. H. Fremlin.

Just put $\mathcal{E}=\Sigma$.

By the way Fremlin's 5 volumes is a must have encyclopedia on measure theory. It contains a hundred times more information on measure theory that you ever want to know.