A single uniform variable supplies countably many independent base-$b$ digits

For $q,k\ge 1$ set $\varphi (q,k)=\frac{(q+k-2)(q+k-1)}{2}+q$. Listing pairs by increasing $q+k$ and then increasing $q$ shows that $\varphi$ is a bijection. Each digit $D_j$ lies in $\{0,\dots ,b-1\}$, and prescribing the first $N$ consecutive digits specifies exactly one half-open interval of length $b^{-N}$ by [F1]. More generally, prescribing any $s$ distinct digit positions among the first $N$ leaves exactly $b^{N-s}$ such intervals, so those digits have joint probability $b^{-s}$.

Fix distinct $q_1,\dots ,q_s$ and positive integers $n_1,\dots ,n_s$. The index sets $\{\varphi (q_i,k):1\le k\le n_i\}$ are disjoint. If their digit values are prescribed, let $N$ be the largest used index. Exactly $b^{N-{\sum }_i{n}_i}$ length-$b^{-N}$ intervals have those values, so their joint probability is $b^{-{\sum }_i{n}_i}$. For each $q$ and $m\ge 1$, the event $T_{q,m}=\{D_{\varphi (q,k)}=b-1\text{ for every }k\ge m\}$ is contained, for every $L\ge 1$, in the event that its first $L$ tail digits all equal $b-1$. Step 1.1 gives $\mathbb{P}(T_{q,m})\le b^{-L}$ for every $L$, so [F5] gives $\mathbb{P}(T_{q,m})=0$. The countable union of these events over $q,m$ is null. Outside it, each stream has no tail consisting entirely of $b-1$ digits; consequently a prescribed initial digit word places its numeric sum in the corresponding half-open base-$b$ interval, not at its right endpoint. Thus every finite family of the numeric ${\tilde{U}}_{q_i}$ has the product uniform cylinder probabilities outside a null set. The event ${\tilde{U}}_q=1$ is contained in $T_{q,1}$, so replacing that value by $0$ leaves every finite-dimensional law unchanged.

Each digit $D_j$ is constant on the finite Borel partition into intervals of length $b^{-j}$. A finite partial sum defining ${\tilde{U}}_q$ is constant on the common refinement at the largest digit index it uses, hence is measurable. Its pointwise limit is measurable by [F4]. The map $g:[0,1]\to [0,1)$ given by $g(x)=x$ for $x<1$ and $g(1)=0$ is Borel; explicitly, for Borel $A\subseteq [0,1)$, $g^{-1}(A)=(A\cap [0,1))$ together with $\{1\}$ exactly when $0\in A$. Therefore $U_q=g({\tilde{U}}_q)$ is measurable. For any finite set of indices $q_i$, the $b$-adic cylinder rectangles for $(U_{q_i})$ form a $\pi$-system generating the Borel sets of $[0,1{)}^s$: every open set is the union of the countably many such cells contained in it, since the cell diameters tend to zero. Step 2.1 gives agreement there with product Lebesgue measure; [F2] therefore gives the product law on all Borel sets. Hence the $U_q$ are mutually independent and each is uniform.

Each map $u\mapsto {\sigma }_{\min \{b!-1,\lfloor b!u\rfloor \}}$ is Borel, with the preimage of ${\sigma }_r$ equal to $[r\mathrm{/}b!,(r+1)\mathrm{/}b!)$ for $0\le r<b!$. For distinct $q_1,\dots ,q_s$, independence and uniformity from step 3.1 give $\mathbb{P}({\pi }_{q_i}={\sigma }_{r_i}\text{ for every }i)={\prod }_i\mathbb{P}(U_{q_i}\in [r_i\mathrm{/}b!,(r_i+1)\mathrm{/}b!))=(b!{)}^{-s}$. These maps are measurable and the joint probabilities show that the ${\pi }_q$ are mutually independent uniform permutations.