From: hsbrand@cs.vu.nl (HS Brandsma)
Newsgroups: sci.math
Subject: Re: Question: Lebesgue Measurable but not Borel
Date: 27 Oct 1997 15:09:02 GMT
Marzocchi Marco (marzocch@ing.unibs.it) wrote:
: Hello,
: I'm looking for an example of a Lebesgue measurable set which is not a
: Borel set. Does anyone can help me?
: Thanks in advance for replies!
:
: Yours,
:
: Marco
A Lebesgue measurable set (in R, say) is just the symmetric difference
of a Borel set and a null-set. So null-sets are where you have to
look: there are 2^continuum many null-sets in R, and only continuum
many Borel sets. So there are loads of examples.
Henno
==============================================================================
From: edgar@math.ohio-state.edu (G. A. Edgar)
Newsgroups: sci.math
Subject: Re: Not a Borel set...please repost
Date: Thu, 30 Oct 1997 10:56:15 -0500
I hope you can decipher the AMS-TeX...
Examples of sets that are (analytic but) not Borel sets:
* $S = \R$ (the real numbers) $T$ is the set of all $x$ with continued fraction
expansion
$$
x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \cdots}}
$$
such that for some positive integers
$r(1) < r(2) < \cdots$, we have
$a_{r(i)}$ divides $a_{r(i+1)}$ for all $i$.
Then $T$ is not a Borel subset of $S$. [Lusin, {\sl Fundamenta Math.}
{\bf 10} (1927) p.~77.]
* $S = \Cal K\big([0, 1]\big)$ (the space of all compact subsets
of the interval $[0,1]$ with the Hausdorff metric), $T \subseteq S$
is the collection
of all uncountable compact sets in $[0, 1]$. The space $S$ is a complete
separable metric space under the Hausdorff metric.
Then $T$ is not a Borel subset of $S$. [Hurewicz, 1930]
* $S = \Cal C\big([0, 1], \R\big)$ (the space of all real-valued
continuous functions on the interval $[0,1]$) and $T$ is the set of
all differentiable functions on $[0, 1]$. The space $S$ is
a complete separable metric space under the metric
of uniform convergence.
Then $T$ is not a Borel subset of $S$. [Mazurkiewicz, 1936]
Additional explicit examples of non-Borel analytic sets are in:
Howard Becker, ``Descriptive set theoretic phenomena in analysis and
topology'', pp.~1--26 in {\it Set Theory of the Continuum}, edited by H. Judah,
W. Just, and H. Woodin, Springer-Verlag, 1992.
--
Gerald A. Edgar edgar@math.ohio-state.edu
==============================================================================
From: mathwft@math.canterbury.ac.nz (Bill Taylor)
Subject: Re: Borel-sets
Date: 4 Mar 1999 04:36:29 GMT
Newsgroups: sci.math
Mike Oliver writes:
|> > :Could someone give me an example of an subset of R (or C) which isn't
|> > :a Borel-set in R (or C).
|>
|> What, you mean you want to take a nonmeasurable set and embed
|> it into a set of measure zero? That's *cheating*;
I'll say! Cheating most foul. :)
|> Hint: Any Borel set is built up from open intervals by
|> iterated operations of complement and countable union. That
|> entire process can be coded into a single real.
|>
|> That gives you a surjection from the reals onto the Borel
|> sets. Diagonalize out.
Hmmm... that's pretty cheaty too! Diagonalizing is most always a bit
codey-logical rather than naturally mathematical. Try the following...
----
This example is due to Lusin (1927). T is the set of all real numbers x
with continued fraction expansion
1
x = a[0] + --------------------------
1
a[1] + ------------------
1
a[2] + ---------
...
such that, for some positive integers r(1) < r(2) < r(3) < ...,
we have a[r(i)] divides a[r(i+1)] for all i.
Lusin showed that this set is not Borel, but is analytic, hence
Lebesgue measurable.
As David Ulrich observed back in 1995, during a previous incarnation
of this thread and this example, the essential reason for its nonborelness
is that it's an uncountable union of borels. Each particular choice
of the r-sequence gives a borel set, and there are uncountably many
(essentially different) of these.
-------------------------------------------------------------------------------
Bill Taylor W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------------------------
Life is a strange attractor in the chaos of the universe.
-------------------------------------------------------------------------------
[Remark: Ullrich states he did not make this claim. --djr]
==============================================================================
In article <36DE139A.9D8DCA8C@math.ucla.edu>,
Mike Oliver wrote:
[Oliver had quoted most of Taylor's article, above through this: --djr]
> As David Ulrich observed back in 1995, during a previous incarnation
> of this thread and this example, the essential reason for its nonborelness
> is that it's an uncountable union of borels.
Hmm? Don't follow that at all. I could give lots of examples of uncountable
unions of borels that are still Borel.
Maybe you mean, uncountably many levels of Borel complexity. Fine, but isn't
that pretty similar to just saying you're coding the countable ordinals into
it?
==============================================================================
From: dlrenfro@gateway.net (Dave L. Renfro)
Subject: Re: nonmeasurable sets and non_Borel sets in R
Date: 27 Aug 1999 11:33:50 -0400
Newsgroups: sci.math
patricia grambsch [sci.math 26 Aug 99 22:29:26 -0400 (EDT)]
asks
>Supposedly, it is difficult but possible to find a subset in R
>constructed explicitly, without the axiom of choice, which
>is not measurable(Lebesgue). Is anyone aware of such a construction
>or a useful reference? There is an uncountable infinity of
>non Borel sets which are Lebesgue measurable.
>Is there an algorithm for explicit construction,
an example, or even an axiom of choice approach?
Under the assumption of a certain large cardinal
hypothesis, it is consistent that the countable
axiom of choice holds (indeed, even the stronger
version known as the Principle of Dependent Choices)
AND that every subset of the reals is Lebesgue
measurable. Thus, you need something "less constructive"
than the countable axiom of choice, which itself
allows for quite a lot. Moreover, your first sentence
is false under any "explicit" notion I know of, but I
believe that the Axiom of Choice (even for just alef_1
many collections of sets of reals) ISN'T a consequence
of the existence of a nonmeasurable set.
If you allow me to make use of ONE nonmeasurable
set, then I can give you uncountably many others
in a simple way.
Suppose E is measurable and non-Borel. Using the
map x |---> tan^(-1)(x), if necessary, we can assume
the set belongs to (-Pi/2, Pi/2). Then any set of the
form
E union A, where A is a measurable subset of [2,3],
is measurable and non-Borel.
[Of course, all but countably many such A's are
not explicit by many uses of "explicit" !]
Note that the construction above gives not just
uncountably many such sets, but 2^c such sets
(c = cardinality of the reals), since there are
2^c measurable (measure zero, in fact) non-Borel
sets in any interval.
You might want to look at the following article.
[I can't find my copy right now, so it may say
something about the things I wasn't entirely sure
of.]
K. Ciesielski, "How good is Lebesgue measure?",
Math. Intelligencer 11(2) (1989), 54-58.
==============================================================================
From: dlrenfro@gateway.net (Dave L. Renfro)
Subject: Re: nonmeasurable sets and non_Borel sets in R
Date: 27 Aug 1999 15:08:17 -0400
Newsgroups: sci.math
Ooops! When I first typed up my response,
I was thinking one of the questions had to do with
an explicit presentation of uncountably many
nonmeasurable sets. Later, I realized the question
actually called for an explicit presentation of uncountably
many measurable non-Borel sets. In making the needed
changes, I forgot to rewrite
>If you allow me to make use of ONE nonmeasurable
>set, then I can give you uncountably many others
>in a simple way.
as
"If you allow me to make use of ONE measurable
non-Borel set, then I can give you uncountably
many others in a simple way."
==============================================================================
From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: nonmeasurable sets and non_Borel sets in R
Date: 27 Aug 1999 16:28:39 -0500
Newsgroups: sci.math
In article ,
patricia grambsch wrote:
>Supposedly, it is difficult but possible to find a subset in R
>constructed explicitly, without the axiom of choice, which
>is not measurable(Lebesgue). Is anyone aware of such a construction
>or a useful reference? There is an uncountable infinity of
>non Borel sets which are Lebesgue measurable.
>Is there an algorithm for explicit construction, an example,
>or even an axiom of choice approach?
The Axiom of Determinateness implies that all sets are
Lebesgue measurable.
There are many ways to use the axiom of choice to get
non-measurable sets. One way is to take all numbers
for which the rational coordinate in a Hamel basis
representation has a particular value.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558