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Source. Published p. 262, Theorem 1.5, and Section 3, p. 272 (PDF). The source gives an outline; the branching inequality and both stopping cases are expanded here, including the fixed-level reduction needed for complements.

Statement. Fix $0<p<1$ and $\eta>0$. There is $c>0$ such that families $\mathcal F,\mathcal G\subseteq2^{[n]}$ avoiding an integer cross intersection $l$ satisfy

$$\mu_p(\mathcal F)\mu_p(\mathcal G)\le e^{-cn} \tag{1}$$

whenever

$$\max(0,2p-1)+\eta\le l/n\le p-\eta. \tag{2}$$

Constants may be chosen uniformly for $p$ in a compact subset of $(0,1)$, with the same positive buffer $\eta$. Thus this gives the source's two open parameter ranges for $l=\lfloor\rho n\rfloor$, after decreasing the buffer and taking $n$ sufficiently large.

Proof. First suppose $p\le1/2$, and put $t=p/(1-p)$ and $c_0=1/t$. Apply the interval deletion algorithm of theorem_1_4, using weighted_deletion_inequality in place of its unweighted step. With fixed sufficiently small $\delta$, each growth step gains $1+\delta$ and each widening step retains $b_{p,\delta}=1-c_0\delta-O_p(\delta^2)$. The interval invariant, termination, and positivity are unchanged.

Let $u,v$ be the two step counts and $m$ the final ambient size. Suppose the initial product is at least $e^{-\delta^2n}$. As in the unweighted proof, logarithms of the gains and the final upper bound one imply

$$u-c_0v\le C_p\delta n,\qquad v\ge p(n-m)-C_p\delta n. \tag{3}$$

Here and below the constants can be enlarged without changing the notation. The final product is at least $e^{-C'_p\delta n}$. These estimates follow from Taylor bounds on a compact parameter interval; all constants are uniform when $p$ is bounded away from zero.

If the procedure stops at $a=0$, its forbidden width is $v$ and $u+v\ge l\ge\eta n$. Hence (3) gives $v\ge p\eta n/2$ for small enough $\delta$. All final cross intersections exceed $v$. Theorem 3.1 and the entropy bound give a final product at most $e^{-v^2/m}\le e^{-p^2\eta^2n/4}$, contradicting the lower bound when $\delta$ is small. Zero ambient size is already impossible for positive families avoiding intersection zero.

If it stops at $b=m$ and $a>0$, then $m=b\le l\le(p-\eta)n$. Using $a=m-v$ and (3),

$$a\le(1+p)m-pn+C_p\delta n \le pm-\eta n/2. \tag{4}$$

Furthermore $a>0$ in the first expression implies $m\ge p n/(2(1+p))$ for sufficiently small $\delta$. All cross intersections are less than $a$. Apply the biased small-intersection bound in product_measure_separation to the actual $m$ coordinates, with the fixed proportional gap supplied by (4). It gives an upper bound $e^{-c_1n}$, where $c_1>0$ depends only on $p,\eta$. Taking $C'_p\delta<c_1$ is a contradiction. This proves (1) for $p\le1/2$ and large $n$.

Now let $p>1/2$. Fix $0<\tau<\eta/8$. Under $\mu_p$, the total measure of sets whose size is outside $[pn-\tau n,pn+\tau n]$ is at most $2e^{-2\tau^2n}$, by the proved concentration estimate. If a family has measure at most twice this quantity, (1) already holds. Otherwise its typical-size part retains at least half its measure. Choose one size $k$ in that part of $\mathcal F$, and one size $h$ in that part of $\mathcal G$, each carrying at least $1/(n+1)$ of its part's measure.

Complement these two uniform families. For their members,

$$|F^c\cap G^c|=n-k-h+|F\cap G|.$$

They therefore avoid $l'=n-k-h+l$. By (2) and $|k-pn|,|h-pn|\le\tau n$,

$$\eta/2\le l'/n\le(1-p)-\eta/2.$$

Their $\mu_{1-p}$ measures equal the original level measures exactly. The already proved case at bias $1-p$ bounds their product by $e^{-c_2n}$. Since the original product is at most $4(n+1)^2$ times that level product, it too has a fixed exponential gap. This proves the second range. Taking common compact bounds on $p$ and $1-p$ proves the asserted uniformity.

For the finitely many remaining $n$, each admissible $l$ is realized by the full cube and every point has positive product measure. Thus no avoiding pair has measure product one. There are finitely many pairs and integers $l$; compactness of the allowed biases, or a smaller constant for a fixed bias, includes these cases. $\square$

Precision. Complementation of arbitrary nonuniform families does not send a fixed intersection to a fixed intersection. The typical-level argument above supplies that missing step. The weighted widening factor has first-order loss $(1-p)\delta/p$, not the unweighted loss $\delta$; its stopping estimate is correspondingly (3).

Dependencies. weighted_deletion_inequality, theorem_1_4, theorem_3_1, product_measure_separation, entropy_estimates.