Timeline for Approximation of Borel sets

8 events

when toggle format	what		by	license	comment
Dec 13, 2017 at 9:25	history	edited	Francesco Polizzi	CC BY-SA 3.0	deleted 1 character in body
Dec 13, 2017 at 8:01	comment	added	Nirav		@JochenWengenroth, Ah I see. Thank you.
Dec 13, 2017 at 7:49	comment	added	Nate Eldredge		@JochenWengenroth: And indeed, if you take some finite weighting of Lebesgue measure, the boundary of a singleton has measure zero, but the boundary of $\mathbb{Q}^2$ doesn't. So the premise of this question is false.
Dec 13, 2017 at 7:48	comment	added	Jochen Wengenroth		The boundary of a countable union is not contained in the union of the boundaries. Write $\mathbb Q^2$ as a union of singletons.
Dec 13, 2017 at 7:28	history	edited	Nirav	CC BY-SA 3.0	fixed grammar
Dec 13, 2017 at 6:37	history	edited	Nirav	CC BY-SA 3.0	included definition of boundary set, included proof of sigma algebra
Dec 13, 2017 at 3:53	comment	added	fedja		*It is not hard to show that $G$ is a $\sigma$-algebra on $\mathbb R^2$*. Really? Have you tried it? (I assume that $\partial A$ is the topological boundary of $A$, if it is not, you'd better tell us what it is before more sarcastic questions come your way).
Dec 13, 2017 at 2:15	history	asked	Nirav	CC BY-SA 3.0