Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a set, and $\cal F$ a family of subsets of $X$, let $\Sigma(\cal F)$ denote the smallest $\sigma$-algebra containing $\cal F$. We can also define $\Sigma(\cal F)$ internally using a transfinite induction up to $\omega_1$. We will say that $\cal F$ is a generating family of $\Sigma(\cal F)$.

For a $\sigma$-algebra $\cal B$, let $\frak g(\cal B)=\min\lbrace|\cal F|:\Sigma(\cal F)= B\rbrace$. Of course this may depend on the space $X$, so let us assume that always $X\in\cal B$, which then can be recovered as $X=\bigcup\cal B$.

It is not hard to see that the $\cal B(\Bbb R)$, the Borel $\sigma$-algebra of $\mathbb R$, has a countable generating family, namely open intervals of the form $(p,q)$ where $p,q\in\mathbb Q$. That is ${\frak g(\cal B(\Bbb R))}=\aleph_0$.

Question I: (ZFC) Let $\cal L$ denote the Lebesgue measurable $\sigma$-algebra. Can we calculate the exact value of $\frak g(\cal L)$ (as a function of $\frak c$)? Is it $2^\frak c$?

Some observations include the following:

  1. Since $|\Sigma(\cal F)|=|\cal F|^{\aleph_0}$ it is trivial that $\frak g(\cal L)>\frak c$, since ${\frak c}^{\aleph_0}=\frak c$.

  2. By the above we have that under GCH (or at least $2^\frak c=c^+$) we can indeed prove the value is equal to $\frak c^+$.

  3. Plain cardinality arguments need not work, suppose $\aleph_1=\frak c$ and $\aleph_{\omega+1}=2^\frak c$ (with SCH). Take $\aleph_\omega$ many subsets, they generate $\aleph_\omega^{\operatorname{cf}(\aleph_\omega)}=2^\frak c$ many subsets. So less than $2^\frak c$ many sets may generate something of the proper size.

  4. Since the Cantor set is Lebesgue measurable of measure zero, its power set is embedded into $\cal L$, so $\frak g(\cal L)\geq\frak g(\cal P(\Bbb R))$, but on the other hand $\cal L$ is a sub-algebra of $\cal P(\Bbb R)$ so we have to have equality.

Question II: (ZFC) If this value is independent of ZFC, does the assertion $\aleph_2=\frak g(\cal L)$ have interesting consequences (e.g. CH, some combinatorial principle, etc.)? Does the assertion $\frak g(\cal L)=\kappa$ imply $2^\mu=2^\frak c$ for all $\frak c\leq\mu\leq\kappa$?

And a bonus question, can we prove anything on this cardinal in Solovay's model, or in $L(\Bbb R)$ when $L(\Bbb R)\models AD$? Is it even well-defined (i.e. is there a minimal size)? Do note that in both these settings Dependent Choice holds. In a broad ZF context if $\Bbb R$ is a countable union of countable sets then the Borel sets cover $\cal P(\Bbb R)$, but there is also a difficulty establishing measure theory properly.

share|improve this question
Concerning the bonus questions: If you want to construct the sigma-algebra by transfinite induction up to $\omega_1$, you need a bit of AC. –  Goldstern Jul 11 '12 at 23:01
Goldstern, more than DC? I mean, the whole point of the construction is that by $\omega_1$ you already finished everything, so DC should be enough. Either way in $L(\Bbb R)$ DC holds when it is a model of AD (or when it is Solovay's model). –  Asaf Karagila Jul 11 '12 at 23:08

1 Answer 1

Concerning question II: We have $2^{\mathfrak c} \le \mathfrak g(\mathcal L) ^{\aleph_0}$.

If CH fails, then $\aleph_2 ^{\aleph_0} = \mathfrak c$, so $\mathfrak g(\mathcal L) = \aleph_2$ is impossible.

Hence $\mathfrak g(\mathcal L) = \aleph_2$ implies CH.

In fact, it is equivalent to $2^{\mathfrak c} = \aleph_2$ or "GCH for $\aleph_0$ and $\aleph_1$".

share|improve this answer
Can that last fact be pushed further? Something in the spirit of MA, where $2^\kappa=\frak c$ for $\kappa<\frak c$, can now be made into $2^\kappa=2^\frak c$ where $\kappa<\frak g(\cal L)$, or something similar... –  Asaf Karagila Jul 11 '12 at 23:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.