Are the open sets of $\{0, 1\}^I$ measurable? Let $I$ be an uncountable set and let $2^I=\{0, 1\}^I$ be the set of all functions from $I$ into $\{0, 1\}$. Consider the product measure $\mu$ on $2^I$. The domain of this measure is the $\sigma$-algebra $\mathcal S$ generated by the clopen subsets of $2^I$ (which coincide with the $\sigma$-algebra generated by the compact $G_\delta$ subsets of $2^I$ - I believe these are known as Baire sets). $\mu$ is the unique measure on $\mathcal S$ such that for every finite subset $F \in I$ and for every $s:F\rightarrow 2$, $\mu(\{f \in 2^I: f|F=s\})=\frac{1}{2^{|s|}}$.
Let $\bar \mu$ be the completion of $\mu$. My question is: are the open subsets of $2^I$ $\bar \mu$-measurable?
I tried to show that if $U$ is open, then the interior and exterior measures of $U$ coincide. I have a pretty good picture of how to work with the interior measure of $U$: $U$ is a union of clopen sets, and it is easy to see that there is a countable union of clopen sets $U'\subseteq U$ such that $\mu(U')=\mu_*(U)$. I don't know how to proceed.
 A: Possible argument that may help.
For the product $\sigma$-field, not every subset of $\{0,1\}^I$ is measurable. Indeed, measurable subsets of $\{0,1\}^I$ can depend only on a countably many coordinates. To see this, for every $D \subset I$, call $\pi_D$ the canonical projection from $\{0,1\}^I$ to $\{0,1\}^D$. Call $\mathcal{S}$ the collection of all subsets $\pi_D^{-1}(B)$, where $D$ is a countable subset of $I$ and $B$ a measurable subset of $\{0,1\}^D$. One checks that $\mathcal{S}$ is a $\sigma$-field and that the canonical projections $\pi_{\{i\}}$ for $i \in I$ are $\mathcal{S}$-measurable. Hence $\mathcal{S}$ contains the product $\sigma$-field of $\{0,1\}^I$.
My impression is that the answer is still negative if you complete the product $\sigma$-field with regard to $\mu$. Indeed, every measurable subset of $\{0,1\}^I$ is almost surely (for the completion of $\mu$) equal to some $\pi_D^{-1}(B)$, where $D$ is a countable subset of $I$ and $B$ a measurable subset of $\{0,1\}^D$.
A: Ok, I think I managed, the answer is positive.
First, some preliminaries.
For each $J\subseteq I$, let $\pi_J:2^I\rightarrow 2^J$ be the natural projection and $\mu_J$ be the product measure on $2^J$ in the product algebra, which coincides with the $\sigma$-algebra $\mathcal B_J$ generated by the clopen subsets of $2^J$. Notice that if $J$ is countable, $2^I$ is second countable, thus this coincides with the Borel subsets of $2^J$.
It is not difficult to see that for every $J\subseteq I$ and for every $\mu_J$-measurable subset $Z$ of $2^J$, $\mu_J(Z)=\mu(\pi_J^{-1}[Z])$. To see that, let $\mathcal B'=\{Z \in \mathcal B_J: \pi_J^{-1}[Z] \text{ is } \mu\text{-measurable and }\mu_J(Z)=\mu(\pi_J^{-1}[Z])\}$. $\mathcal B'$ clearly contains the basic clopen sets (which are sets of the form $\{f \in 2^J: f|K=s\}$ where $K\subseteq J$ is some finite set and $s:K\rightarrow 2$). $\mathcal B'$ is also clearly closed by intersections, set differences and countable disjoint unions, so it contains the $\sigma$-algebra generated by the basic clopen sets, which contain the clopen sets, concluding the proof of the desired equality.
Moreover, suppose $J\subseteq I$, $A\subseteq 2^J$ is $\mu_J$-measurable and $B\subseteq 2^{I\setminus J}$ is $\mu_{I\setminus J}$-measurable. Then $\mu(\pi_J^{-1}[A]\cap\pi_{I\setminus J}^{-1}[B])=\mu_J(A).\mu_{I\setminus J}(B)$. To see this, fix a basic clopen $A$ and prove that this holds for all $\mu_{I\setminus J}$-measurable $B\subseteq 2^{I\setminus J}$, then fix such a $B$ and show that the expression holds for all intended $A$'s. The details are easily filled.
Now we proceed to the proof.
Let $W\subseteq 2^I$ be an open set. By the topology of $2^I$ is zero-dimensional, there exists a collection $\mathcal W$ of basic clopen sets such that $\bigcup \mathcal W= W$.
Now we exhaust $\mathcal W$, that is, obtain a countable set $\mathcal W'\subseteq \mathcal W$ such that for every $U \in \mathcal W$, $\mu(U\setminus \bigcup \mathcal W')=0$. To see that such a set exists, let $c=\sup\{\mu(\bigcup \mathcal V):\mathcal V \subseteq W \text{ is countable}\}$, let $\mathcal V_n$ be a sequence of countable subsets of $\mathcal W$ such that $\mu(\mathcal V_n)\rightarrow c$ and let $\mathcal U'=\bigcup_{n \in \mathbb N} \mathcal V_n$. Then $\mu(\mathcal U')=c$, which clearly implies by the maximality of $c$ that $\mathcal U$ has the desired propery.
Let $W'=\bigcup \mathcal W'$. $W'$ is open. By the definition of basic open set, for each $U \in \mathcal W'$ there exists a finite $K_U\subseteq I$ and $s_U:K_u\rightarrow 2$ such that $U=\{f \in 2^I: F|K_U=s_u\}$. Let $J=\bigcup_{U \in \mathcal W'}K_U$. Then $J$ is countable.
Let $V=\pi_J[W]$ and $V'=\pi_J[W']$. Since $W, W'$ are both open subsets of $2^J$ and $J$ is countable, they both are $\mu_J$-measurable. Notice that $\pi_J^{-1}[V']=\pi_J^{-1}[\pi_J[W']=W'$. The last equality holds by the choice of $J$.
I claim that $\mu_J(V)=\mu_J(V')$. Suppose not. Notice that $V=\bigcup_{U \in \mathcal W}\pi_J[U]$. Since $2^J$ is hereditarelly Lindelof, there exists a countable  $\mathcal W''\subseteq \mathcal W$ such that $V=\bigcup_{U \in \mathcal W''}\pi_J[U]$. It follows that $\mu_J(\bigcup_{U \in \mathcal W''}\pi_J[U]\setminus V')>0$, so there exists $U \in \mathcal W''$ such that $\mu_J(\pi_J[U]\setminus V')>0$. Fix $U$.  We will get a contradiction by showing that $\mu(U\setminus W')>0$.
Recall $U=\{f \in 2^I: f|K_U=s_U\}$. Let $K=K_U$, $s=s_U$. By using that $\pi_{J}^{-1}[\pi_J[W']]=W'$. Notice that $\pi_J[U\setminus W']=\pi_J[U\setminus\pi_J^{-1}[\pi_J[W']]]=\pi_J[U]\setminus \pi_J[W']=\pi_J[U]\setminus V'$, which is $\mu_J$-measurable. Likewise, $\pi_{I\setminus J}[U]$ is a basic clopen set, thus is measurable. With all this in mind, it is easy to show that $U\setminus W'=\pi_J^{-1}[\pi_J[U\setminus W']]\cap \pi_{I\setminus J}^{-1}[\pi_{I\setminus J}[U]]$. Thus, $\mu(U\setminus W')=\mu_J(\pi_J[U]\setminus V').\mu_{I\setminus J}(\pi_{I\setminus J}[U])$, and both measures are stricly positive.
Finally, since $\mu_J(V)=\mu_J(V')$, we have that $\mu(W')=\mu(\pi_J^{-1}[\pi_J[W]])$. Now notice that $W'\subseteq W\subseteq \pi_J^{-1}[\pi_J[W]]$, which completes the proof since $W=W'\cup N$, where $W'$ is measurable and $N$ is contained in the null set $\pi_J^{-1}[\pi_J[W]]\setminus W'$.
