Re: support of a finite borel measure

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 08/03/04 

Date: Tue, 03 Aug 2004 08:50:32 -0500

On Tue, 03 Aug 2004 08:23:37 -0400, "G. A. Edgar"
<edgar@math.ohio-state.edu.invalid> wrote:

>In article <0fbtg0hbmht8s444ucqfmdj6d6jeco2dso@4ax.com>, David C.
>Ullrich <ullrich@math.okstate.edu> wrote:
>
>> i never recall when one can show a measure is regular [because
>> where i come from it doesn't come up...]. if you can show that
>> mu is inner regular you're done, since any compact subset of
>> G is covered by finitely many open null sets.
>
>He just said separable, not complete. So you cannot expect there
>to exist lots of compact sets.

i suppose not. never mind...

>An example would be a Bernstein set
>in R: a set E such that both E and its complement meet every
>uncountable compact set in R.
>
>So E is a separable metric space. Lebesgue outer measure on [0,1],
>restricted to the Borel sets of E, give you a finite probability
>measure. But for that measure, all compact sets are null,
>because all compact sets in E are countable.

************************

David C. Ullrich

sorry about the inelegant formatting - typing
one-handed for a few weeks...

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