Cardinality of measurable sets

Question

It is known that the cardinality of the set of all Lebesgue measurable sets in $[0,1]$ is $2^\mathfrak c$. I am asking what will be the cardinality of the set of measurable sets $[0,1]^I$, where $I$ is any cardinal.

comment: Can I compute it like $(2^\mathfrak c)^I=2^{\mathfrak c\cdot I}$

What will correspond the Cantor set here ?

Answer 1

If by measurable you mean measurable with respect to product measure (on product of Borel algebras), then the answer is the same as in case of $[0,1]$: there are as many measurable sets as there are all subsets.

The argument for that is mostly the same as in case of the interval, even easier in fact:

1. Fix a coordinate $i\in I$.
2. Then any subset of the $x(i)=0$ fiber, that is to say, any set $X\subseteq [0,1]^I$ such that for all $x\in X$ we have $x(i)=0$ is null (because the fiber is null) and thus measurable.
3. But if $I$ has at least two elements, then clearly the fiber has the same cardinality as the entire product, so there are as many such sets as there are subsets of $[0,1]^I$, i.e. $2^{{\mathfrak c}^{\lvert I\rvert}}$.

The same answer applies if you replace $[0,1]$ by an arbitrary continuous probability space. A slightly more careful argument will, I think, do the trick for any probability space, at least if $I$ has at least two elements.