Measurability problem of sample distribution function of a contiuum of independent random variable Let $I = [0,1]$ be the index set of a contiuum of i.i.d random variables. For each $t \in I$, the sample space of $X_t$ is $\Bbb R$ equipped with Borel $\sigma$-algebra and Borel probability measure. By Kolmogorov extension theorem, we have a unique probability measure on the sample space $\Omega = {\Bbb R}^{I}$.
Specifically, for each Lebesgue measurable subset of $\Bbb R$, $A$, we have $A^t = \{\omega \in \Omega: \omega(t) = A\}$. Then all sets of the form $A_t$ constitute a $\pi$-system that generates the $\sigma$-algebra of $\Omega$, which is, $\mathscr{F}$. The measure $\mu$ of $\Omega$ must be consistent with finite dimentional distributions in the sense that: $$\mu(A^t) = m(A)$$ $$\mu(\bigcap_{1 \leq i \leq n}A^{t_i}) = \prod_{1 \leq i \leq n} \mu(A^{t_i}) $$
The mapping $i \mapsto t_i$ is one-on-one for each $n$.$(\omega, \mathscr{F}, \mu)$ is the desired probability space.
For each $c \in \Bbb R$, the sample distribution function $F_{\omega}(\cdot)$ is defined as: $$F_{\omega}(c) = l(\{ t \in I: X_t(\omega) \leq c\})$$ $l$ is the lesbesgue measure.
In Judd(1985)(Paywall required), in $\sf {ZFC}$, $\omega_c = \{t \in I: X_t(\omega)  \leq c\}$ could be non-measurable, thus $F_{\omega}(c)$ could also be non-measurable.
Here's screenshot of the theorem:



$\mu^{\ast}(N)$ and $\mu_{\ast}(N)$ are outer and inner measures of $N$ respectively.
Question: What does it mean "any Borel set is restricted on at most a countable number of indices" in the Fifth line of the proof?
 A: A general fact of measure theory is that if $X$ is a set and $\mathcal{F}$ a family of subsets and $B\in\sigma(\mathcal{F})$, then there exists a countable subfamily $\mathcal{C}\subseteq\mathcal{F}$ such that $B\in\sigma(\mathcal{C})$. In order to prove this, one just has to verify that the sets with this property form a $\sigma$-algebra containing all elements of $\mathcal{F}$. Applied to product spaces, if a set is in the product-$\sigma$-algebra, then it is generated by countably many coordinate projections. 
For a modern account of the problem Judd discusses, see the introduction of Podczeck, On existence of rich Fubini extensions (paywall). The problem was known for quite some time before Judd. Doob already mentioned that sample-realizations will usually be non-measurable. In the paper Zur Gleichwertigkeit zweier Arten der Randomisierung from 1974, von Weizsäcker has Judd's proof as a step in the proof of his main theorem.
See also the related discussion on Mathoverflow.
A: We can check that the $\sigma$-algebra generated by the $A_t$ consists of the sets of the form $\{\omega\in\Omega\mid (\omega(t_j),j\geqslant 1)\in B\}$, where $(t_j,j\geqslant 1)$ is a sequence of real numbers, and $B$ a Borel subsets of $\mathbb R^{\infty}$, the set of sequences of real numbers endowed with the product Borel $\sigma$-algebra. 
The main difficulty here is to prove that this collection is closed under countable unions.
