Dear all,
For any sigma-finite measure $\mu$ on the Borel subsets of a metric space X (separable if needed), we recall that a $\mu$-continuity set is any Borel set B such that $\mu(boundary of B) = 0$
I would like to find a good reference (book or article) where is treated clearly the proof of the result which establishes that the sigma-algebra generated by the $\mu$-continuity sets is equal to the Borel sigma-algebra.
Sincerely,
Bogachev, Measure theory, vol 2 Proposition 8.2.8. μ-continuous sets form a sub-algebra of Borel field, which contains the base of topology.
Bogachev, Measure theory, vol 2 Proposition 8.2.8. μ-continuous sets form a sub-algebra of Borel field, which contains the base of topology.
This is a good question.
A paper that considers the relation between mu continuity and Borel sigma algebra is
https://www.researchgate.net/publication/221965801_A_Generic_Property_of_Exact_Magnetic_Lagrangians
See Lemma 6 on page 7.
Article A Generic Property of Exact Magnetic Lagrangians
I too have noticed some difference in terminology; sometimes it depends on what has been read, if its an old article. Generally one of the conditions which allowed for finite measures to be compatible with countably infinite algebras, or to be extendable to a sigma algebra; so to be compatible with even a countably additive measure. I know of some papers, where they suggest that they can achieve compatibility results if not uniqueness results, even ifs its not an algebra (i think it was kopylov 2010, 2005).
Otherwise there has been some confusion in the past, it was once stated is necessary and sufficient, and that required the existence of a countable base in the order topology, perfect separably or a countably - dense subset, although it only requires a countable "dense -order" subset if you luck and krantz, suppes (foundations of measurement vol 1). Or otherwise Jaffray (1974) see attached, Chautenaff(1984,1981), see attached, Fishburne (1986) or fine (1973). If, however, one does not have monotone continuity, and/or, the space is not completely connected (which incidentally does not required that the ordering satisfies between-nness, or that its literally a dense ordering or a denumerable complete desnse ordering, as in the rationals in the reals or something close to atom-less-ness) then one does need it to almost a dense ordering in this sense, or otherwise be perfectly separable plus a whole bunch of archimedean conditions (or something ones strong one such as H).
As far as I can gather, this is both distinct both from the ordering of some part of, or even all of the ordering of algebra being dense as in the rational in the reals, or that one can merely fit an dense set of fictitious numbers, ordered so as to be dense (the inequalities being, connected, by >, =. < and such that that for A>B there exists a C such that A
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