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

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On the other hand, it's established that in non-$$\sigma$$-finite spaces some weird things can happen (Fubini's theorem no longer valid, etc.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.

• these are two different things. separability is determined by topology and has pretty much nothing to do with measure while $\sigma$-finiteness depends on the particular measure. you could for example take your space to be $\mathbb{R}$ - clearly it's separable and then put a counting measure on it (so the measure of a set would be the number of its points) Commented Jun 1, 2014 at 19:28
• @mm-aops I think you're mistaking a separable topological space with a measure and a topological space with separable measure. It seems that you're describing the first option, while I'm looking for the information on the second one. I'll add details to my post to avoid confusion. Commented Jun 1, 2014 at 20:17
• all right, my bad, cheers Commented Jun 1, 2014 at 21:25
• @Norbert why did you delete your answer? Commented Jun 15, 2014 at 13:47
• I don't understand at all your notion of separability of measure. For me a measure space is separable iff the collection of finite $\mu$-measurable sets endowed with your distance form a separable space. If the space is non $\mu$-finite, seems to me that the collection of $\mu$-measurable cannot form a metric space, then with your notion is not clear how to deal with these cases Commented May 31 at 18:18

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.

• that's boring... Commented Jun 16, 2014 at 6:29
• @Norbert Indeed. Commented Jun 16, 2014 at 7:03
• I don't understand why you said that $\Sigma$ is separable. By Norbert definition a measure space is separable iff measurable sets form a separable metric space with the exhibited distance. You're $\Sigma$ does not form a metric space because $d(X, \emptyset) = +\infty$ Commented May 31 at 18:04
• Not Norbert definito Sorry, i mean the definition on the question Commented May 31 at 19:29
• @ManuelBonanno If you want all such values to be finite, there are no infinite measure spaces. Commented May 31 at 23:25

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.

Just put $$\mathcal{E}=\Sigma$$.
• I do not see how this implies separability. In fact, I think we can just choose $\mathcal E_0 = \{X\}$ in (iv).