From: David Ullrich Newsgroups: sci.math Subject: Re: Are the Borel sets recognizable inside the Lebesgue sets? Date: Tue, 15 Oct 1996 12:14:08 -0500 Jonathan King wrote: > > On the unit interval, the field (sigma-algebra) B of Borel sets (the smallest > field containing the open sets) is a proper subfield of L, the Lebesgue > measurable sets. > > Just knowing countable boolean operations/relations (complement, > countable-union, countable-intersection, inclusion) operate on L, and knowing > the measure of each member of L, is it possible to detect when a member is > actually a Borel set? No. Well, I'm not _certain_ I know exactly what the question means, but here's a true fact which indicates the answer's no if the question means what I think it does: True Fact: There exists a bijection of L to L which preserves Lebesgue measure and countable Boolean operations, but which takes at least one Borel set to a non-Borel set. Pf: Let K denote the traditional middle-thirds Cantor set. The measure of K is 0 so every subset of K is Lebesgue measurable, but not every subset of K is Borel, since K has 2^c subsets and there are only c Borel sets. Now in fact we can find a non-Borel set X contained in K such that both X and the relative complement K\X have cardinality c. (For example, the left half of K has 2^c subsets of cardinalilty c; any non-Borel one of those will do.) We can also find a Borel set Y contained in K such that Y and K\Y both have cardinality c (for example the left half of K). Now there exists a bijection from K to K that takes X to Y and Y to X; fix one such, extend it to a bijection of the unit interval to itself by declaring it to be constant on the complement of K, and call it f. Define a bijection from L to L by composition with f. (Note it's not true that a measurable function of a measurable set must be measurable, but it's easy to see that _this_ f actually maps L to L.) -- David Ullrich ?his ?s ?avid ?llrich's ?ig ?ile (Someone undeleted it for me...)