Lebesgue Measurable but not Borel sets.

[Bacle2]

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 2^c=2^2Aleph_0 such subsets.

iii)*** The process of producing the Fσ , G_δ , F_σδ ,...

produces only 2^Aleph_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.

 

[micromass]

This is a bit complicated.
For (i), you are basically asking why there are only 2^{\aleph_0} 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

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

If \alpha is an ordinal and if \mathcal{A}_\alpha is defined, then define \mathcal{A}_{\alpha+1} such that

• If A\in \mathcal{A}_\alpha, then \mathbb{R}\setminus A\in \mathcal{A}_{\alpha+1}.
• If A_n\in \mathcal{A}_\alpha for n\in \mathbb{N}, then \bigcup_n{A_n}\in \mathcal{A}_{\alpha+1}

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

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

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

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

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

For (ii), there is this explicit construction of a non-Borel Lebesgue set: http://planetmath.org/ALebesgueMeasurableButNonBorelSet.html

 

[Bacle2]

I see; it's been a while since I saw this. Thanks.

 

[Bacle2]

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.