Why isn't $[0,1]^{\aleph_1}$ isomorphic to $[0,1]$ as a measure space.

Question

I was talking with a professor, and he mentioned that $[0,1]$ (with Lebesgue $\sigma$-algebra and Lebesgue measure) isn't isomorphic as a measure space to $[0,1]^{\aleph_1}$ (with the product measure). It is intuitive, but I could not prove this, and i haven't been able to find a reference. Does anyone know how to prove it or where to find a proof?

Answer 1

Expanding on what I wrote in the comments:

The reason we need $\aleph_1$ is that $[0, 1]^{\aleph_0}$ is isomorphic to $[0, 1]$. This has to do with the fact that $[0, 1]$ is isomorphic to $\{0, 1\}^{\aleph_0}$ as a measure space, where $\{0, 1\}$ has the uniform probability measure (think about binary expansions). Then $(\{0, 1\}^{\aleph_0})^{\aleph_0}$ is isomorphic to $[0, 1]$, because $\aleph_0 \times \aleph_0 = \aleph_0$. This comes up for instance in the proof of lemma 12 here and there's some related discussion here.

In my mind this is quite closely related to the fact that one can define a countably infinite sequence of iid random variables on $[0, 1]$ which are uniformly distributed in $[0, 1]$. The proof I learned of this was via "Rademacher functions" - essentially, turning each $x \in [0, 1]$ into an infinite binary sequence, splitting that up into infinitely many sequences, and stitching those back together to obtain infinitely many different real numbers. It's clear that this proof won't show that there are uncountably many such variables, and indeed we can argue that that's not possible. Such an argument is given here, and it relies on the separability of $L^2([0, 1])$.

Since there is an obvious $\aleph_1$-sized family of iid random variables on $[0, 1]^{\aleph_1}$ given by the coordinate projections, we are done. Here the point is that a measure space isomorphism of $X$ and $Y$ induces a normed-vector-space isomorphism of $L^p(X)$ and $L^p(Y)$.

Indeed we can see slightly more directly that $L^p([0, 1]^{\aleph_1})$ is not separable, without worrying too much about all the probability concepts, by considering the coordinate projections. If $\pi_i$ and $\pi_j$ are two distinct projections, then $\lVert \pi_i - \pi_j \rVert_p$ is bounded below by some positive constant. For example, let $A$ be the set $\{\omega \in [0, 1]^{\aleph_1} : \omega_i \in [0, \tfrac 13], \omega_j \in [\tfrac 23, 1]\}$. Then $A$ has positive measure ($\tfrac 19$), and on $A$, we have $|\pi_i - \pi_j| \ge \tfrac 13$, so $\lVert \pi_i - \pi_j \rVert_p \ge (\tfrac 19)^{1/p}(\tfrac 13)$. (You can also actually work out the integral if you like but that's not why I learned measure theory :)). This shows $L^p([0, 1]^{\aleph_1})$ is not separable as we have found an uncountable family of functions which are all far apart.

It's well known that $L^p([0, 1])$ is separable for $p < \infty$. For instance, the set of "rational-endpoint step functions" is dense. So we're done!