Lebesgue Measurable but not Borel sets.

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  • #1
Bacle2
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Hi, All:

I am trying to find a construction of a measurable subset that is not Borel, and ask

for a ref. in this argument ( see the ***) used to show the existence of such sets:

i) Every set of outer measure 0 is measurable, since:

0=m* (S)≥m*(S) , forcing equality.

ii) Every subset of the Cantor set is measurable, by i), and there are 2c=22Aleph_0 such subsets.

iii)*** The process of producing the Fσ , Gδ , Fσδ ,.....

produces only 2Aleph_0 sets. ***

iv) Since the 2 cardinalities are different, there must be a set as described in ii), i.e., a Lebesgue-measurable set that is not Borel.

So, questions:

1)How do we show the process in iii) produces only c sets.

2)Anyone know of an actual construction of this set?

Thanks.
 

Answers and Replies

  • #2
22,089
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This is a bit complicated.
For (i), you are basically asking why there are only [itex]2^{\aleph_0}[/itex] Borel sets. This is proved in a lot of axiomatic set theory books.

Let me type up the basic idea. You can always ask for more details if you want. Define

[tex]\mathcal{A}_1=\{B\subseteq \mathbb{R}~\vert~\text{B is open}\}[/tex]

If [itex]\alpha[/itex] is an ordinal and if [itex]\mathcal{A}_\alpha[/itex] is defined, then define [itex]\mathcal{A}_{\alpha+1}[/itex] such that
  • If [itex]A\in \mathcal{A}_\alpha[/itex], then [itex]\mathbb{R}\setminus A\in \mathcal{A}_{\alpha+1}[/itex].
  • If [itex]A_n\in \mathcal{A}_\alpha[/itex] for [itex]n\in \mathbb{N}[/itex], then [itex]\bigcup_n{A_n}\in \mathcal{A}_{\alpha+1}[/itex]

If [itex]\gamma[/itex] is a limit ordinal, and [itex]A_\alpha[/itex] is defined for [itex]\alpha<\gamma[/itex], then define
[tex]\mathcal{A}_\gamma=\bigcup_\alpha \mathcal{A}_{\alpha}[/tex]

It is clear that if [itex]\mathcal{B}[/itex] are the Borel sets, then each [itex]\mathcal{A}_\alpha[/itex] is a subset of [itex]\mathcal{B}[/itex].

It can easily be proven by transfinite induction that each [itex]\mathcal{A}_\alpha[/itex] with [itex]\alpha\leq \omega_1[/itex] has only [itex]2^{\aleph_0}[/itex] sets.

We now show that [itex]\mathcal{A}_{\omega_1}=\mathcal{B}[/itex]. For this, it suffices to show that [itex]\mathcal{A}_{\omega_1}[/itex] is a sigma-algebra containing the open sets. This is not hard to show. Let me show, for example that if [itex]A_n\in \mathcal{A}_{\omega_1}[/itex] for [itex]n\in \mathbb{N}[/itex], then [itex]\bigcup_n{A_n}\in \mathcal{A}_{\omega_1}[/itex].

Indeed, if [itex]A_n\in \mathcal{A}_{\omega_1}[/itex], then there exists an ordinal [itex]\alpha_n<\omega_1[/itex] such that [itex]A_n\in \mathcal{A}_{\alpha_n}[/itex]. Let [itex]\alpha=\sup_n \alpha_n[/itex]. Since each [itex]\alpha_n[/itex] is countable, it follows that [itex]\alpha[/itex] (as a union of the [itex]\alpha_n[/itex]) is countable as well. So [itex]\alpha<\omega_1[/itex].
Now, since [itex]A_n\in \mathcal{A}_{\alpha_n}\subseteq \mathcal{A}_\alpha[/itex] for each n, it follows by definition that [itex]\bigcup_n A_n \in \mathcal{A}_{\alpha+1}[/itex]. But since [itex]\alpha<\omega_1[/itex], we also know that [itex]\alpha+1<\omega_1[/itex]. So we deduce that [itex]\bigcup_n A_n\in \mathcal{A}_{\alpha+1}\subseteq \mathcal{A}_{\omega_1}[/itex].

For (ii), there is this explicit construction of a non-Borel Lebesgue set: http://planetmath.org/ALebesgueMeasurableButNonBorelSet.html [Broken]
 
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  • #3
Bacle2
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I see; it's been a while since I saw this. Thanks.
 
Last edited:
  • #4
Bacle2
Science Advisor
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Hope it is not too OT, but I wonder if you have a ref. too, for how to

construct models of the reals ( or, more generally, of 1st-order theory in the standard structure of the reals,

for other structures. ) of different cardinalities, re Lowenheim-Skolem. I would like to

avoid forcing if possible. If not, I will post it somewhere else, sorry.

Thanks.

Seriously sorry for bothering you so much; just to tell you I posted this in 'Set Theory ...' forum, so feel free to delete my post if necessary.
 
Last edited:

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