lark
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No. There are examples of Borel measures on [itex]\sigma[/itex]-compact (even on compact) spaces which fail to be regular.And I suppose a Borel measure on a [itex]\sigma[/itex]-compact space like the complex plane has to be regular then?
[itex]\mu[/itex] is associated with a bounded linear functional by [itex]\Phi(f) = \int_Xfd\mu[/itex]. Then by the Riesz representation theorem, if X is locally compact & Hausdorff, [itex]\Phi[/itex] is associated with a regular measure [itex]\mu^\prime[/itex] by [itex]\Phi(f) = \int_Xfd\mu^\prime[/itex]. So [itex]\int_Xfd(\mu-\mu^\prime)=0,[/itex] all [itex]f[/itex] in [itex]C_0(X)[/itex].morphism said:Can you explain how you're "regularizing" [itex]\mu[/itex]?
If X is locally compact and Hausdorff can this still happen? example?No. There are examples of Borel measures on [itex]\sigma[/itex]-compact (even on compact) spaces which fail to be regular.
The standard example is X=[0,w] where w is the first uncountable ordinal. This is an exercise in Rudin (last one in chapter 2 if you have the first edition).lark said:If X is locally compact and Hausdorff can this still happen? example?
morphism said:As for your other question, I don't really know what happens in general. I'll think about it some more and let you know if I come up with anything.