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Replace $\mathcal{A}$ by its upward closure of a set family $\overline{\mathcal{A}}$; this remains an intersecting family and can only make the desired containment easier. Apply the regularity lemma for Boolean functions to its indicator with parameters $(\epsilon /10,p,r,\epsilon /2)$, where $r$ is chosen from part (ii) with density threshold $\epsilon /2$. We obtain a bounded set $J$ such that all but $\epsilon /2$ of the ${\mu }_p$-weighted restrictions are $(\epsilon /10,p,r)$-quasirandom.

Let $\mathcal{B}\subseteq \mathcal{P}(J)$ consist of assignments $u$ for which the restriction is quasirandom and has $p$-biased expectation at least $\epsilon /2$. Restrictions excluded because of irregularity contribute at most $\epsilon /2$, and the remaining excluded restrictions contribute at most $\epsilon /2$ by their conditional density. Hence

It remains to prove that $\mathcal{B}$ is intersecting. If disjoint $u,v\in \mathcal{B}$ existed, part (ii) would give ${\mathbb{E}}_{1/2}{f}_u>1/2$ and ${\mathbb{E}}_{1/2}{f}_v>1/2$. Couple two unbiased complementary assignments on $[n]\setminus J$. Since two subsets of a common finite probability space having measures greater than $1/2$ must intersect, some complementary pair would make both restrictions equal to one. Together with disjoint $u,v$, this would produce two disjoint members of $\mathcal{A}$, a contradiction. Therefore $\mathcal{B}$ is intersecting, proving the Dinur-Friedgut junta theorem for intersecting families with $m=|J|$.

Solved by gpt-5.6-sol high.

A Boolean function $f:\{0,1{\}}^n\to \{0,1\}$ is $(\epsilon ,p,r)$-quasirandom when, for every $J\subseteq [n]$ with $|J|\le r$ and every $u\in \{0,1{\}}^J$,

The regularity lemma for Boolean functions states that for every $\epsilon ,p,r,\delta$ there is $T$ such that every Boolean function has a set $J$, $|J|\le T$, for which a ${\mu }_p$-random $u\in \{0,1{\}}^J$ satisfiesHere $f_u$ is the restriction obtained by fixing the coordinates in $J$ to $u$.

Solved by gpt-5.6-sol high.