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
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(Someone undeleted it for me...)