Forming a subset of $\mathbb{R}$ by coin tossing

I assert that the answer is undefined.

Let's describe $(\Omega,\Sigma,\Pr)$, the probability space of $2^{\aleph_0}$ coins being tossed, one for each real number. The "Borel" sets in $\Sigma$ are the $\sigma$-algebra generated by sets of the form $\Pi_{r \in \mathbb R} F_r$ where finitely many $F_r$ are not $\emptyset$ or $\{H,T\}$, and the probability of this set of $0$ if any $F_r = \emptyset$, and otherwise $\frac12$ to the power of the number of $r$ such that $F_r$ has exactly one element in it. Then $\Sigma$ is the completion of the "Borel" sets, by adding in subsets of "Borel" sets of measure zero.

Let $$ B = \{ w : X(w) \text{ is measurable} \} .$$ I assert that $B$ is not measurable.

We say that $A \subset \Omega$ depends upon $Y \subset \mathbb R$ if for every $(c_r) \in \Omega$, we have that $(c_r) \in A$ is completely determined by the values of $c_r$ for $r \in Y$. We can see that all Borel subsets of $\Omega$ depend upon a countable subset of $\mathbb R$. So we can see that $B$ isn't a "Borel" set.

Lemma: If $B \subset A$ and $A$ is "Borel," then $A = \Omega$.

Suppose $(c_r) \in \Omega \setminus A$. Suppose $A$ depends upon $Y \subset \mathbb R$ where $Y$ is countable. Let $$ Z = \{r \in Y : c_r = H \} .$$ Then unwinding the definitions, we see that there exists a countable $Y$ and $Z \subset Y$ such that for all $X \subset \mathbb R$, that $X \cap Y = Z$ implies $X$ is not measurable. But clearly this is not true. $\square$

Corollary: The outer measure of $B$ is 1.

Exactly the same argument can be applied to $\Omega \setminus B$, so:

Lemma: The inner measure of $B$ is 0.

So $B$ is non-measurable in the worst possible way. And $\Pr(B)$ is undefined.

Note 1: You could ask far more mundane questions like, "What is the probability that $X$ has measure $0$?" Answer: undefined.

Note 2: I put Borel in quotes. True Borel sets are the sigma algebra created by open sets in the product topology, which is not quite what I have. For example, the subset of sequences $(c_r)$ such that at $c_r = T$ for at least one $r \in \mathbb R$ is open in the product topology, but not "Borel" using the definition I have given. I think that my "Borel sets" are equivalent to Baire sets (the $\sigma$-algebra generated by close $G_\delta$ sets, also defined as the minimal $\sigma$-algebra such that continuous functions from $\Omega \to \mathbb R$ are continuous.)