This question assumes the existence of a large real-valued measurable cardinal. Let $X$ be an uncountable set and $(X,2^X)$ equipped with a non-atomic probability distribution $P$. Additionally, let $Q$ be the "uniform" distribution on the set $\{0,1\}^X$ --- i.e., the product of independent Bernoulli(1/2) random variables indexed by $X$ --- whose existence follows from the Łomnicki-Ulam theorem. The $\sigma$-algebra for the latter probability space is the usual cylindrical one used for random processes.

Let us now draw a "random function" $h\sim Q$ and a random point $x\sim P$.

Question: Is the object $h(x)$ a measurable random variable on the corresponding product space?

  • $\begingroup$ Do you mean that $X$ carries a nonatomic total measure? $\endgroup$ – Monroe Eskew Jan 9 at 13:58
  • $\begingroup$ Yes, a nonatomic total probability measure. $\endgroup$ – Aryeh Kontorovich Jan 9 at 14:03
  • $\begingroup$ I think that the fact that $X$ is real-valued-measurable doesn't help. I think the measure space you're describing is just the Lebesgue measure on binary $X$-sequences. $\endgroup$ – Monroe Eskew Jan 9 at 14:15
  • $\begingroup$ So we can do Vitali's argument to show there's a nonmeasurable set and thus $h$ might not be a measurable function. Let $\{ x_n : n \in \mathbb N \}$ be a sequence of distinct elements of $X$. For binary $X$-sequences $f,g$, say $f \sim g$ iff there is a finite $s \subseteq \mathbb N$ such that $f$ is the result of flipping the bits of $g$ on $\{ x_n : n \in s \}$. Each $s$ induces a measure-preserving transformation, corresponding to translation in $\mathbb R$. Each equivalence class is countable, and you know the rest... $\endgroup$ – Monroe Eskew Jan 9 at 14:29
  • $\begingroup$ @MonroeEskew I think your description of "Lebesgue measure on binary sequences" applies to the uniform measure on {0,1}^X with its cylindrical sigma-algebra but not to (X,2^X). I've updated the question to clarify. $\endgroup$ – Aryeh Kontorovich Jan 9 at 14:42

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