A positive triple set has uniformly overlapping sections

Put $\eta ={\varphi }^{\otimes 3}(E)>0$. The product measure is finite because $\varphi$ is finite. Fix a generating sequence $(G_j{)}_{j\ge 1}$ for $\mathcal{X}$. Let ${\mathcal{P}}_n$ consist of the nonempty sets ${\bigcap }_{j=1}^n{H}_j$, where each $H_j$ is $G_j$ or $G_j^c$. These are finite partitions and ${\mathcal{P}}_{n+1}$ refines ${\mathcal{P}}_n$. Their finite-union algebras increase and contain every $G_j$. The union of the corresponding tensor-cell algebras on ${\mathsf{X}}^3$ is therefore an algebra generating ${\mathcal{X}}^{\otimes 3}$: it contains every coordinate lift of a $G_j$, which generate the product sigma-algebra, and all its cells are product-measurable. Apply [F2] to obtain a member $H$ of that algebra with ${\varphi }^{\otimes 3}(E\mathrm{△}H)<\eta \mathrm{/}6$. Then ${\varphi }^{\otimes 3}(H)>5\eta \mathrm{/}6$ and ${\varphi }^{\otimes 3}(H\setminus E)<\eta \mathrm{/}6<{\varphi }^{\otimes 3}(H)\mathrm{/}4$, so $E$ occupies more than three quarters of $H$. The set $H$ is a finite union of tensor cells at a common partition level. If every positive-measure cell had occupancy at most three quarters, finite additivity would give the same upper bound on $H$, a contradiction. Thus some cell $U\times V\times W$ has positive product measure and ${\varphi }^{\otimes 3}(E\cap (U\times V\times W))>\frac{3}{4}\varphi (U)\varphi (V)\varphi (W)>0$. Only one finite-algebra approximant and one cell of its finite partition are selected; no conditional-expectation representative or choice axiom is used.

Since $E\subseteq A{\times }_{\text{\hspace{0.17em}⁣}y}B$, Tonelli gives ${\int }_{U\times V}{1}_A(x,y)d\varphi (x)d\varphi (y)>\frac{3}{4}\varphi (U)\varphi (V),$ and likewise ${\int }_{V\times W}{1}_B(y,z)d\varphi (y)d\varphi (z)>\frac{3}{4}\varphi (V)\varphi (W)$. Consequently the sets $C=\{x\in U:\varphi (A_1(x)\cap V)>\frac{3}{4}\varphi (V)\}$ and $D=\{z\in W:\varphi (B_2(z)\cap V)>\frac{3}{4}\varphi (V)\}$ have positive measure; otherwise the corresponding integral could not exceed three quarters of the rectangle measure.

For $x\in C$ and $z\in D$, both sections occupy more than three quarters of $V$. The complement of their intersection in $V$ is the union of two sets, each of measure less than $\varphi (V)\mathrm{/}4$. Thus $\varphi (A_1(x)\cap B_2(z))\ge \varphi (V)\mathrm{/}2>0$. Set $\gamma =\varphi (V)\mathrm{/}2$.