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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.

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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 $\aleph_1$".

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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

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