# Lebesgue Measurable but not Borel sets.

1. Jul 30, 2012

### 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 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.

2. Jul 31, 2012

### micromass

Staff Emeritus
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 [Broken]

Last edited by a moderator: May 6, 2017
3. Jul 31, 2012

### Bacle2

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

Last edited: Jul 31, 2012
4. Jul 31, 2012

### 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.

Last edited: Jul 31, 2012