22 апр. 2010 г. · Let X be a metric space. Then every Borel measure μ on X is regular (i.e. for every Borel set B and every ε > 0, there exists a closed set Fε ...

13 окт. 2012 г. · A measure μ on Rn is Borel regular if μ is Borel and for each A⊂Rn there exists a Borel set B such that A⊂B and μ(A)=μ(B). A measure μ on Rn ...

21 апр. 2013 г. · A measure is regular if for which every measurable set can be approximated from above by an open measurable set and from below by a compact measurable set.

4 авг. 2019 г. · The statement you are looking for is probably that every locally finite Borel measure on a separable complete metric space X is regular.

5 окт. 2018 г. · Let S be a compact and μ a regular Borel measure on S with total variation 1. If for every partition of S into two Borel subsets the measure of ...

4 апр. 2012 г. · Does an atom of such measure have to be a singleton (up to set of zero measure)?

20 мар. 2011 г. · PS: A Borel measure ν on Rn is regular if ν(K)<∞ for every compact K. From this definition we have ν(E)=inf{ν(U)|Uopen,U⊇E}. A signed or complex ...

29 окт. 2016 г. · I attempted to prove that any finite measure is regular. However, I did not see a clear reason for introducing the family R.

26 июл. 2016 г. · Yes, any regular Borel outer measure m is topologically additive. Indeed, take any S,T⊆X such that S⊆U and T⊆V for some disjoint open ...

11 окт. 2019 г. · A finite Borel measure on X is regular if and only if it is outer regular on all Borel sets and inner regular on all open sets.