How to prove that there can't exist a countable set $\{A\_1,A\_2,\dots\}\subset \mathcal{L}(\mathbb{R})$ (where $\mathcal{L}(\mathbb{R})$ denotes the family of all Lebesgue measurable sets) such that $\sigma(\{A\_1,A\_2,\dots\})=\mathcal{L}(\mathbb{R})$?


(I know that the cardinality of $\mathcal{L}$ is equal to the cardinality of all subsets of $\mathbb{R}$ and hence larger then the cardinality of the borel sets but I'm not familiar with any deeper hierarchy results. Is there a more or less elementary proof?)



How to prove that the Lebesgue $\sigma$-Algebra is not countably generated?
Let me mount the kind of cardinality argument to which you allude.


You had asked for a proof that the $\sigma$-algebra of Lebesgue measurable sets is not countably generated. But in fact, a much stronger claim is true:


**Theorem.** The $\sigma$-algebra of all Lebesgue measurable sets is not generated by any family of size continuum.


**Proof.** First, I claim that the $\sigma$-algebra generated by any family of size at most continuum itself has size at most continuum. (In particular, every countably generated $\sigma$-algebras have size at most continuum.)


The reason is that we can perform the generation process concretely in a transfinite process of length $\omega\_1$. We begin with the generating family, which has size at most continuum. At each stage, we form the next collection by taking all possible countable unions, countable intersections, and complements of what we already have; at limit stages of the process, we gather together all the sets we have formed before that stage. If we began with at most continuum many sets, then since there are only at most continuum many countable subsets of a set of size continuum, it follows that the next stage will also have size at most continuum. And at countable limit stages, we have a countable union of previous families, all of size at most continuum, and so the collection at that stage also has size at most continuum.


By stage $\omega\_1$, we have generated the $\sigma$-algebra, since all countable collections from what we have at that stage have already appeared altogether at a countable stage. So we have $\omega\_1$ many stages, each with at most size continuum many sets, making the generated family of size at most continuum, which establishes the initial claim.


Second, since every subset of a Lebesgue measure zero set is measurable, and there are sets, such as the Cantor set, which have measure zero and size continuum, it follows that there are $2^{2^{\aleph\_0}}$ many Lebesgue measurable sets. This is strictly larger than the continuum, and so by the above it cannot be generated by any family of size at most continuum. $\Box$


