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Jan 14 at 16:37 history edited Daniele Tampieri CC BY-SA 4.0
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Feb 23, 2017 at 18:01 comment added Julian Newman As this is an important question I just want to summarise the key points from the different answers below: $(X,\mathcal{S},\mu)$ is a measure space. (I) The semi-metric $\mu(A \triangle B)$ is separable if and only if $\exists$ countable set $\mathcal{S}_0 \subset \mathcal{S}$ s.t. $\mathcal{S} \subset \sigma(\mathcal{S}_0 \cup \mathcal{N}_\mu)$. (II) It is possible that the semi-metric $\mu(A \triangle B)$ is separable but there is no countable set $\mathcal{S}_0 \subset \mathcal{S}$ s.t. $\mathcal{S} \subset \sigma(\mathcal{S}_0 \cup \mathcal{N}_{\mu|_{\sigma(\mathcal{S}_0})})$.
May 28, 2012 at 20:46 answer added Vaughn Climenhaga timeline score: 9
Jun 9, 2010 at 8:40 vote accept Kestutis Cesnavicius
Jun 9, 2010 at 2:31 comment added Joel David Hamkins Kestutis, I explain how to do this in my updated answer. You can use the finite Boolean combinations of the generating set.
Jun 8, 2010 at 21:07 comment added Kestutis Cesnavicius Say the $\sigma$-algebra is countably generated. How do I pick the countable dense subset in the semi-metric given by $\mu(A\Delta B)$?
Jun 8, 2010 at 19:25 answer added coudy timeline score: 10
Jun 8, 2010 at 19:19 answer added Joel David Hamkins timeline score: 15
Jun 8, 2010 at 19:16 comment added Nate Eldredge I'm too lazy to try to prove this right now, but where does your proof get stuck?
Jun 8, 2010 at 18:43 comment added François G. Dorais For convenience, I copied the Wikipedia article in a community wiki answer.
Jun 8, 2010 at 18:42 answer added François G. Dorais timeline score: 4
Jun 8, 2010 at 18:18 history asked Kestutis Cesnavicius CC BY-SA 2.5