Skip to main content
8 events
when toggle format what by license comment
Feb 17, 2020 at 12:46 comment added user95282 One necessary condition is that the measure must be perfect; or, if we are dealing with not necessarily finite measure spaces, the restriction to every set of finite measure must be perfect. For a completion of a countably generated space with a finite measure this is also sufficient. For more on perfect measures, see Chapter 52 in Fremlin's Measure Theory.
Feb 14, 2020 at 21:01 comment added Piotr Hajlasz @JohannesHahn OK. I misunderstood the question. I was not a careful reader. I am deleting my stupid comments.
Feb 14, 2020 at 20:10 comment added Johannes Hahn @PiotrHajlasz How do get $C_0(X)$, if you only know $(X,\Sigma,\mu)$ ? In particular: How do you get the topology from the measure? In general you can't, since the topology is not uniquely determined from the measure space and there are measure spaces that do not come from a topological space at all. So how do you decide whether you're in such a case? And if you're not, how do you decide if one of the possible compatible topologies is LCH ?
Feb 14, 2020 at 19:46 comment added Johannes Hahn @PiotrHajlasz I think the question is about the "if you have a measure on a LCH space" part. How do you know that you are in that situation? If I were to give you any measure space, how would you decide whether or not the sigma algebra is the borel-sets of some LCH space? And can you decide it in a way that gives you enough information to see if my measure is Radon or not?
Feb 14, 2020 at 16:52 comment added Nate Eldredge The "bible" of measure theory is Fremlin's book. It's dense but you could look there.
Feb 14, 2020 at 16:46 comment added Nate Eldredge I guess for starters, you need some condition for a measurable space to be the Borel $\sigma$-algebra of some LCH topology. I would be interested in that in itself.
Feb 14, 2020 at 16:27 history edited YCor CC BY-SA 4.0
edited tags
Feb 14, 2020 at 15:54 history asked Rupert CC BY-SA 4.0