Talk:Universally measurable set

Finiteness condition

So it seems I goofed in my first definition; looking around, everyone seems to impose some finiteness condition on the measure. Does this actually matter? Can someone cook up a universally measurable set of reals that's not measurable with respect to, say, Hausdorff measure of dimension 1/2 ? --Trovatore 16:27, 3 October 2005 (UTC)[reply]

You asked me to comment on this. Unfortunately, my set theoretic knowledge is rather limited. In particular, I don't know what Polish spaces or analytic sets are. So, can't help much. Oleg Alexandrov 03:55, 4 October 2005 (UTC)[reply]

Some such A ?

In the section "Example contrasting with Lebesgue measureability", it says that "Thus we can think of A as a subset of the interval [0,1], and evaluate its Lebesgue measure". Later is says "there are some such A without a well defined Lebesgue measure". It is unclear whether the "some such A" in the second statement refers to a set not in 2^omega or whether the first statement should have said "attempt to evaluate its Lebesgue measure". Tashiro (talk) 15:57, 7 October 2012 (UTC)[reply]

I made an edit following your suggestion. Thank you. 185.69.144.34 (talk) 18:22, 4 September 2018 (UTC)[reply]

Missing properties

Hello, I suggest properties of universally measurable sets should be added. Such that whether intersections and unions (countable?) are universally measurable, and continuous images and preimages. Thank you 148.252.128.209 (talk) 08:26, 19 May 2018 (UTC)[reply]