The original law, fine archive and retained information

The recursions are $u_j=2u_{j-1}-s_j$ and $v_j=(v_{j-1}+s_j)/2$. Induction gives $u_n=\{2^nu_0\}$ and the displayed expression for $v_n$. The integer $\lfloor2^nu_0\rfloor$ has binary expansion $\sum_j2^{n-j}s_j$, while the integer in the archive has expansion $\sum_j2^{j-1}s_j$. This proves the reversal. To invert, put $k=\lfloor Mb\rfloor$:

For example, $u_0=3/10,v_0=2/5,n=2$ gives $(r,b)=(1/5,3/5)$. Omitting reversal would give $b=7/20$, an incorrect archive. Each inverse branch has determinant one; the ideal map is a measure-preserving bijection modulo its null seams.

On an interval $I=[a,b]$ of length $\ell$, its average $f_I$ obeys

This proof includes BV jumps, by the one-dimensional variation measure. Apply it to the $M$ writer cells at fixed $(v,N)$ and let $P_Mf$ be their cell averages. Then $\|f-P_Mf\|_1\leq V_u/(2M)$.

On the final archive strip $k=\lfloor Mb\rfloor$, the output density is $f((r+\operatorname{rev}_n k)/M,Mb-k,N)$. Integrating in $r$ is exactly averaging $f$ over the corresponding original writer cell. Consequently $B^n_\#((P_Mf)\lambda)=dr\,\nu_n$. Measure preservation and the half-variation convention give (4.2). Bit reversal only permutes the writer cells and does not remove or average the final archive.

Every prefix preserves the complete reference; its pushforward density remains at most $C$. Couple the physical and ideal iterations by the same original point. Until their first entry outside $L_h\cup R_h$ the states and the next complete maps agree exactly. A union bound using the ideal reference-preserving prefixes charges at most $nC\beta_h$ for the first failure. Their endpoint coupling proves (4.3). The same good event identifies each actual pre-stage declaration with the ideal one and proves the digit assertion.

Insert the ideal law and its actual nuisance marginal between the two laws in (4.4). Lemma 4.2 gives the middle fee. The endpoint fee is $nC\beta_h$, and marginalizing that comparison and tensoring with $dr$ costs another at most $nC\beta_h$. This is why the own-marginal freshness bound contains a factor two.

For (4.5), apply the same argument to $\mu_B$ and its own nuisance submarginal. The outside-box output and $dr$ times its actual nuisance marginal have equal mass $\tau$, so their TV is at most $\tau$. Adding these two subprobability comparisons proves the result without recutting the law at a later time.

The maps are invertible and reference preserving (modulo null seams for the ideal baker). Their pushforward density is $f\circ F^{-1}$. Changing variables proves equality of the integrals of $|f-1|/2$ and $f\log f$; the negative part of $f\log f$ is bounded on a probability reference, so the extended integral is well defined. The conditional-freshness target is $dr$ times the actual nuisance marginal, not the original full reference $\lambda$. There is therefore no contradiction.

One-dimensional BV slicing gives $g(u)\leq A_i(u_{-i})$ almost everywhere, where $A_i$ is the integral plus variation of its $i$th slice. Thus $g^{3/2}\leq(A_1A_2A_3)^{1/2}$. Integrate first in $u_1$ and apply Cauchy to $A_2^{1/2}A_3^{1/2}$. Then apply Cauchy in $(u_2,u_3)$ to the remaining $A_1^{1/2}$ and the separated product. This gives $\int g^{3/2}\leq\prod_i(\int A_i)^{1/2} =\prod_i(m+V_i)^{1/2}$. Taking the $2/3$ power and arithmetic-geometric mean proves (4.7). Hölder with conjugate exponent three proves the event estimate.

Every complete ideal prefix preserves the $L^{3/2}$ norm. Its next exceptional set has the same volume $\beta_h$, so Hölder bounds its boxed mass by the conditional concentration coefficient times $\beta_h^{1/3}$. Average using the actual $X$ law and sum over $n$ first-failure events. The boxed map fee is at most $nB\beta_h^{1/3}$. The proof of Theorem 4.3, including its own-marginal factor two and one original tail, then gives (4.8). For the displayed row, $\beta_h^{1/3}<2\,10^{-36}$ and the map fee is below $5.2663\,10^{-11}$; the dyadic term and tail give the stated loose ceiling.