Lebesgue but not Borel
Stephen J. Herschkorn
but I have never seen a description of a Lebesgue set of reals which is
not a Borel set. Anyone have an example? Can it be done in ZF?
It suffices to describe a (necessarily uncountable) subset of a null
Borel set which itself is not Borel.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
David Bernier
Stephen J. Herschkorn wrote:
> I am familiar with the construction of a non-measurable set of reals,
> but I have never seen a description of a Lebesgue set of reals which is
> not a Borel set. Anyone have an example? Can it be done in ZF?
>
> It suffices to describe a (necessarily uncountable) subset of a null
> Borel set which itself is not Borel.
The non constructive approach I've seen starts with the Cantor set,
which has cardinality continuum and has Lebesgue measure 0.
As it happens, the cardinality of the Borel sets is only
the continuum. However, the Cantor set has 2^continuum subsets,
all of measure 0, and since 2^continuum> continuum, there
must be subsets of the Cantor set with Lebesgue measure 0
which are not Borel sets. I've never seen any explicit
example...
David Bernier
Jyrki Lahtonen
Let f:[0,1]->[0,1] be the Cantor function (i.e. the thing that's constant
on each interval in the complement X of the Cantor set C). It is continuous and
increasing, so g(x)=f(x)+x is a continuous and strictly increasing function
from [0,1] to [0,2]. Therefore g is a homeomorphism. The graph of g has slope
exactly one on each interval in X, so the measure of g(X) is equal to 1.
Therefore g(C) has also measure 1. We know that there exists a non-measurable
subset Y of g(C). Consider the set Z=g^{-1}(Y). It is a subset of C, hence
has outer measure zero, hence is measurable. However, Z cannot be a Borel
set, for then Y=g(Z) would also be a Borel set (a homeomorphism maps a Borel set
to another).
Cheers,
Jyrki Lahtonen, Turku, Finland
Mikko Kangasmaki
: I am familiar with the construction of a non-measurable set of reals,
: not a Borel set. Anyone have an example? Can it be done in ZF?
Take 1-dimensional non-measurable set and consider it as a subset of 2-dimensional plane.
Bacause it is subset of x-axis, it must have measure zero, so it is measurable and
therefor a Lebesgue set. On the other hand it cannot be Borel-set, because otherwise the
original (1-dimensional) set would be measurable.
Yhis reqires axiom of choice, because the nonmeasurable set is constructed with it.
(But I dont think that its such a bad thing because axiom of choise is used all the time
without even knowing it.)
m.k.
G. A. Edgar
<hers...@rutcor.rutgers.edu> wrote:
> I am familiar with the construction of a non-measurable set of reals,
> but I have never seen a description of a Lebesgue set of reals which is
> not a Borel set. Anyone have an example? Can it be done in ZF?
>
> It suffices to describe a (necessarily uncountable) subset of a null
> Borel set which itself is not Borel.
Yes, it can be done in ZF. One can explicitly construct an analytic
set whose complement is not analytic. Such a set is Lebesgue
measurable
but not Borel.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Fred Galvin
> I am familiar with the construction of a non-measurable set of reals,
> but I have never seen a description of a Lebesgue set of reals which is
> not a Borel set. Anyone have an example? Can it be done in ZF?
Yes, it can be done in ZF. Here is an outline of a proof. Let R be the
set of all real numbers, Q the rational numbers, I the irrational
numbers. Let C be the Cantor set. Let B be the set of all Borel sets
of real numbers.
1. Define a surjection f from R to B. (You can do this without AC. You
would need AC to get a bijection, but you don't need a bijection.)
2. Let X = {x in R: x is not in f(x)}. Note that X is not a Borel set.
3. Let Y = X \ Q. Note that Y is not a Borel set.
4. Define a homeomorphic embedding h of I into C.
5. Note that h(Y) is not a Borel set and has Lebesgue measure 0.