0
$\begingroup$

It is known that the Borel $\sigma$-algebra ${\cal B}(\mathbb{R}) $ is countably generated. And it is also known that ${\cal B}(\mathbb{R}) \subsetneq {\cal L}(\mu^*, \mathbb{R}), $ the $\sigma$-algebra of Lebesgue measurable sets. In fact, the latter is the completion of the former. Is there a simple way to prove that ${\cal L}(\mu*, \mathbb{R}) $ is not countably generated?

In the beginning I thought that since $ {\cal L}(\mu^*, \mathbb{R}) $ contains ${\cal B}(\mathbb{R}) $ and the closure, a countable collection cannot generate it, unless the closure itself is countable, which one should prove it exists. I also wondered if the fact that $ {\cal L}(\mu^*, \mathbb{R}) $ contains all singletons as its only atoms, could be exploited, but I ended up with some headache instead. Does anyone have a good suggestion?

Thank you.

Maurice.

Improve this question
$\endgroup$
1
$\begingroup$

One way is to prove that any countably generated $\sigma$-algebra has cardinality at most that of the continuum. That takes some time to show, but it is done in the top answer to this question Is the intersection of two countably generated $\sigma$-algebras countably generated?. Then recall that any subset of the Cantor set is Lebesgue measurable so there are clearly more than $\aleph_1$ of them.

Btw, you shouldn’t equate countable generability and separability: the Lebesque sets are separable in the metric induced by the measure.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Thank you. I was just thinking about the $\sigma$-algebra being or not being countably generated which some authors call separable and not the sets themselves. $\endgroup$ – Maurice Apr 3 '20 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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