[proofplan]
We use the standard complete-marker lemma for aperiodic automorphisms to choose a small complete section whose first $N$ iterates are disjoint. The first-return decomposition over this marker partitions almost all of the space into Rokhlin columns. In each return column, we group levels into consecutive blocks of length $N$ and discard only the marker-level remainder, whose total measure is controlled by the smallness of the marker. The union of the bases of these length-$N$ blocks gives the desired tower.
[/proofplan]
[step:Choose a small marker set with disjoint first $N$ iterates]
Fix $N\in\mathbb{N}$ and $\varepsilon>0$. We use the following standard complete-marker lemma for aperiodic invertible probability-preserving transformations on standard probability spaces: for every $n\in\mathbb N$ and every $\eta>0$, there is a measurable set $M\in\mathcal A$ such that
\begin{align*}
0<\mu(M)<\eta,
\end{align*}
the sets
\begin{align*}
M,TM,\dots,T^{n-1}M
\end{align*}
are pairwise disjoint, and the two-sided saturation of $M$ has full measure:
\begin{align*}
\mu\left(X\setminus \bigcup_{j\in\mathbb{Z}}T^jM\right)=0.
\end{align*}
This marker lemma is the usual complete-section selection theorem; it is the only external measurable-selection input in the proof. Applying it with $n=N$ and $\eta=\varepsilon/N$, choose $A:=M$. Then $A\in\mathcal A$ satisfies
\begin{align*}
0<\mu(A)<\frac{\varepsilon}{N},
\end{align*}
the sets
\begin{align*}
A,TA,\dots,T^{N-1}A
\end{align*}
are pairwise disjoint, and
\begin{align*}
\mu\left(X\setminus \bigcup_{j\in\mathbb{Z}}T^jA\right)=0.
\end{align*}
Invertibility is used only to speak about the two-sided saturation, and measure preservation keeps all iterates measurable with the same measure.
[guided]
The proof needs a marker set that is both small and complete. The precise input is the standard complete-marker lemma for aperiodic invertible probability-preserving transformations on standard probability spaces: given $n\in\mathbb N$ and $\eta>0$, one can find $M\in\mathcal A$ with
\begin{align*}
0<\mu(M)<\eta,
\end{align*}
such that
\begin{align*}
M,TM,\dots,T^{n-1}M
\end{align*}
are pairwise disjoint and
\begin{align*}
\mu\left(X\setminus\bigcup_{j\in\mathbb Z}T^jM\right)=0.
\end{align*}
This is a complete-section statement: the set $M$ meets almost every two-sided orbit, but it has no return during the first $n$ iterates. It is not obtained by saturating a marker after the fact; the completeness and the first-return gap are part of the same lemma.
Apply the lemma with $n=N$ and $\eta=\varepsilon/N$, and set $A:=M$. Then
\begin{align*}
0<\mu(A)<\frac{\varepsilon}{N},
\end{align*}
the levels
\begin{align*}
A,TA,\dots,T^{N-1}A
\end{align*}
are pairwise disjoint, and almost every point of $X$ belongs to a two-sided iterate of $A$:
\begin{align*}
\mu\left(X\setminus\bigcup_{j\in\mathbb Z}T^jA\right)=0.
\end{align*}
These are exactly the three marker properties used in the return-column construction below.
[/guided]
[/step]
[step:Decompose almost all points into first-return columns over the marker]
Define the first-return time to the marker
\begin{align*}
r_A: A \to \mathbb{N}\cup\{\infty\}
\end{align*}
by letting $r_A(x)$ be the least integer $m\ge 1$ such that $T^m x\in A$, and setting $r_A(x)=\infty$ if no such integer exists. By the Poincare Recurrence Theorem, applied to the positive-measure set $A$, we have $r_A(x)<\infty$ for $\mu$-almost every $x\in A$.
For each integer $m\ge 1$, define the measurable return base
\begin{align*}
A_m:=\{x\in A:r_A(x)=m\}.
\end{align*}
The sets $(A_m)_{m=1}^{\infty}$ partition $A$ up to a null set. Since the first $N$ iterates of $A$ are pairwise disjoint, $A_m$ is null for $1\le m<N$. For each $m\ge N$, the sets
\begin{align*}
A_m,TA_m,\dots,T^{m-1}A_m
\end{align*}
are pairwise disjoint, because no point of $A_m$ returns to $A$ before time $m$.
The saturation assumption and recurrence imply that the return columns cover $X$ up to a null set:
\begin{align*}
\mu\left(X\setminus \bigcup_{m=N}^{\infty}\bigcup_{j=0}^{m-1}T^jA_m\right)=0.
\end{align*}
[guided]
The marker set $A$ replaces the invalid attempt to find a block inside the complement of a nearly maximal tower. We deliberately choose $A$ with small measure and no self-intersection during the first $N$ iterates; the smallness will control the final discarded remainder, while the disjointness forces every return column to have height at least $N$.
Define
\begin{align*}
r_A: A \to \mathbb{N}\cup\{\infty\}
\end{align*}
by taking $r_A(x)$ to be the least integer $m\ge 1$ satisfying $T^m x\in A$, and by setting $r_A(x)=\infty$ if no such integer exists. The set $A$ is measurable and has positive measure. Since $T$ is probability-preserving, the Poincare Recurrence Theorem applies and gives $r_A(x)<\infty$ for $\mu$-almost every $x\in A$.
For each $m\ge 1$, define
\begin{align*}
A_m:=\{x\in A:r_A(x)=m\}.
\end{align*}
This set is measurable because it is the finite intersection of $A$, $T^{-m}A$, and the complements $X\setminus T^{-j}A$ for $1\le j<m$. The sets $(A_m)_{m=1}^{\infty}$ partition $A$ modulo the null set on which $r_A=\infty$.
Why are the short return bases null? If $1\le m<N$ and $x\in A_m$, then $x\in A$ and $T^m x\in A$, so $x\in A\cap T^{-m}A$. This contradicts the pairwise disjointness of $A,TA,\dots,T^{N-1}A$ except on a null set. Hence $A_m$ is null for $1\le m<N$.
For $m\ge N$, the column
\begin{align*}
A_m,TA_m,\dots,T^{m-1}A_m
\end{align*}
is pairwise disjoint. Indeed, if two levels $T^iA_m$ and $T^jA_m$ with $0\le i<j<m$ met on a positive-measure set, then applying $T^{-i}$ would give a positive-measure subset of $A_m$ returning to $A$ at time $j-i<m$, contradicting the definition of $r_A=m$.
Finally, because the two-sided saturation of $A$ has full measure and almost every point of $A$ returns to $A$, almost every point of $X$ lies between two successive visits to $A$. Therefore the return columns cover $X$ up to a null set:
\begin{align*}
\mu\left(X\setminus \bigcup_{m=N}^{\infty}\bigcup_{j=0}^{m-1}T^jA_m\right)=0.
\end{align*}
[/guided]
[/step]
[step:Group each return column into blocks of length $N$]
For each $m\ge N$, write
\begin{align*}
m=q_mN+s_m
\end{align*}
with integers $q_m\ge 1$ and $0\le s_m<N$. Define the block base inside the $m$-th return column by
\begin{align*}
B_m:=\bigcup_{\ell=0}^{q_m-1}T^{\ell N}A_m.
\end{align*}
Then define
\begin{align*}
B:=\bigcup_{m=N}^{\infty}B_m.
\end{align*}
The set $B$ is measurable. Since each return column is pairwise disjoint and the columns are disjoint from one another up to null sets, the levels
\begin{align*}
B,TB,\dots,T^{N-1}B
\end{align*}
are pairwise disjoint up to null sets. Removing a null subset of $B$ if necessary, we may assume they are pairwise disjoint as measurable sets.
The levels missed in the $m$-th column are exactly the final $s_m$ levels
\begin{align*}
T^{q_mN}A_m,T^{q_mN+1}A_m,\dots,T^{m-1}A_m.
\end{align*}
Their total measure is $s_m\mu(A_m)\le (N-1)\mu(A_m)$. Summing over $m\ge N$ and using disjointness of the partition of $A$ gives
\begin{align*}
\mu\left(X\setminus \bigcup_{i=0}^{N-1}T^iB\right)
\le
\sum_{m=N}^{\infty}s_m\mu(A_m)
\le
(N-1)\sum_{m=N}^{\infty}\mu(A_m)
\le
N\mu(A)
<\varepsilon.
\end{align*}
[/step]
[step:Conclude that the constructed tower has small remainder]
Set
\begin{align*}
U:=\bigcup_{i=0}^{N-1}T^iB.
\end{align*}
The previous step proves
\begin{align*}
\mu(X\setminus U)<\varepsilon.
\end{align*}
The set $B\in\mathcal{A}$ has pairwise disjoint levels $B,TB,\dots,T^{N-1}B$ by construction. Therefore $(B,TB,\dots,T^{N-1}B)$ is a Rokhlin tower of height $N$ whose remainder has measure less than $\varepsilon$, as required.
[/step]
