Borel's exceptional set can be uncountable and null

Let $C_n$ require only digits $2,4,\dots ,2n$ to be zero. Prescribing the first $2n$ digits leaves $n$ odd-position digits free, so [F1] writes $C_n$ as a disjoint union of $2^n$ half-open cylinders, each of length $2^{-2n}$. Consequently $\lambda (C_n)=2^{-n}$.

For $A\subseteq {\mathbb{N}}_{>0}$, form the binary string whose digit at position $2r-1$ is $1$ exactly when $r\in A$, and whose even-position digits are all $0$. Its series $x_A={\sum }_{r\in A}{2}^{-(2r-1)}$ lies in $[0,1)$. After any place its tail has a forced zero, so the tail value is strictly less than $1$; the greedy recurrence therefore recovers exactly this string. Thus $A\mapsto x_A$ maps $\mathcal{P}({\mathbb{N}}_{>0})$ into $C$.

Every string in $C$ has every even digit zero, so two adjacent digits can never both equal one. The word $11$ has frequency $0$, not the required $1\mathrm{/}4$; hence no member of $C$ is normal in base two.

The sets $C_n$ decrease and $C={\bigcap }_{n\ge 1}{C}_n$. Since $\lambda (C_1)<\mathrm{\infty }$, continuity from above gives $\lambda (C)={\inf }_n{2}^{-n}=0$.

Different subsets have different first differing odd digit, so canonical uniqueness makes the map injective. Conversely, the set of odd positions at which a point of $C$ has digit one recovers it, so the map is a bijection onto $C$.

If $C$ were countable, [F3] would give a surjection $\mathbb{N}\to C$. Composing it with the inverse bijection in step 2.2 and the explicit shift bijection between $\mathbb{N}$ and ${\mathbb{N}}_{>0}$ would give a surjection ${\mathbb{N}}_{>0}\to \mathcal{P}({\mathbb{N}}_{>0})$, contradicting Cantor's theorem. Hence $C$ is uncountable.

Steps 1.3, 2.1, and 3.1 give an uncountable null subset of the exceptional set in [F4]. Countable choice is inherited from the Lebesgue/cylinder suppliers; the family and the coding map are explicit.