Talk:Uniqueness of Measures

The proof as given by Schilling uses Dynkin systems. It is essential that $\mathcal G$ is closed under finite intersections, by Dynkin System with Generator Closed under Intersection is Sigma-Algebra. The cover is indeed a valid replacement for the exhausting sequence; that is an exercise in Schilling I haven't put up yet (but with the generalisation in place, the reference can just be added). --Lord_Farin 06:21, 25 March 2012 (EDT)

You're right that $\mathcal G$ needs to be closed under finite intersections (which, using mathematical induction, can be weakened to intersection of two elements, as I'm sure you know). Thanks for pointing that out, and sorry for overlooking that. I'll add that requirement to the theorem. –Abcxyz (talk | contribs) 10:49, 25 March 2012 (EDT)

Could the 'General Statement' be replaced by, say:

'Alternatively, the exhausting sequence in $(2)$ may be replaced by a countable $\mathcal G$-cover $\left({G_n}\right)_{n \in \N}$'

I would say that such results in a cleaner statement. --Lord_Farin 10:58, 25 March 2012 (EDT)

Sure, but of course all elements of the countable cover must have finite measure. I don't know if that is a clearer (or cleaner) way of stating this; you probably have better judgment than I do when it comes to these things. –Abcxyz (talk | contribs) 11:05, 25 March 2012 (EDT)

That work for you? --Lord_Farin 11:08, 25 March 2012 (EDT)

Just (please) don't remove any hypotheses, of course. Other than that, I really don't have any other opinions (as for now). –Abcxyz (talk | contribs) 11:15, 25 March 2012 (EDT)