Borel's normal number theorem

Fix $b\ge 2$, $\mathrm{\ell }\ge 1$, and $w\in \{0,\dots ,b-1{\}}^{\mathrm{\ell }}$. Put $m={\sum }_{r=1}^{\mathrm{\ell }}{w}_r{b}^{\mathrm{\ell }-r}$, $I_w=[m\mathrm{/}{b}^{\mathrm{\ell }},(m+1)\mathrm{/}{b}^{\mathrm{\ell }})$, and $f_w={\mathbf{1}}_{I_w}$. Then $0\le f_w\le 1$ and $\int f_{w}d\lambda =b^{-\mathrm{\ell }}$.

By [F1] and [F2], $A_n{f}_w$ converges almost everywhere to an invariant function $F_w$, and $F_w=c_w$ almost everywhere for some constant $c_w$. Because $0\le A_n{f}_w\le 1$, [F3] and invariance of the integral give $c_w=\int F_{w}d\lambda ={\lim }_n\int A_n{f}_{w}d\lambda =\int f_{w}d\lambda =b^{-\mathrm{\ell }}\mathrm{.}$

For $x\mathrm{\notin }{E}_b$, [F5] identifies $A_n{f}_w(x)$ with $N_n(w,x)\mathrm{/}n$. Thus outside the union of $E_b$ and the exceptional set from step 2.1, the word $w$ has limiting frequency $b^{-\mathrm{\ell }}$.

The triples $(b,\mathrm{\ell },w)$ form a countable family: $b$ and $\mathrm{\ell }$ range over integers and, for each pair, there are only $b^{\mathrm{\ell }}$ words. By [F6], the union of their null exceptional sets and the countable endpoint sets $E_b$ is null. Every point outside that union satisfies step 3.1 for every base and word, hence is normal. Countable choice is inherited from [F1], [F4], and [F5]; the indexing and cylinders are explicit.