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\begin{fclemma}[]
  % Lemma 7.4
  \label{lemma:7.4}
  For every $\zeta, \alpha > 0$ and $p \in \left[ \zeta, \half - \zeta \right]$,
  there exists $\eps > 0$ and $r$ such that if $f$ is an \quasi{\eps, \rho, r}
  monotone Boolean function from $\{0, 1\}^n$ to $\{0, 1\}$ with $\Ebb f^{(p)} =
  \alpha$, then $\Ebb f^{(1 / 2)} > \half$.
\end{fclemma}

\begin{proof}
  Suppose that $\Ebb f^{(1 / 2)} \le \half$.
  By the Mean Value Theorem, there exists $s \in \left( p, \half \right)$ such
  that
  \[
    \frac{\dd}{\dd s} \Ebb f^{(s)}
    \le \frac{\half - \alpha}{\half - p}
    \le \frac{1}{\zeta}
  .\]
  By the main corollary to the \nameref{thm:6.5} (\cref{coro:6.6}), it follows
  that $\totinf(f^{(s)}) \le \frac{1}{\zeta}$.
  By the \nameref{thm:6.11}, we can find a Boolean $J$-junta $h$ such that
  $\Pbb[f^{(s)} \neq h^{(s)}] \le \eps$ and $|J| \le r(\zeta, \eps)$.

  But $\Ebb f^{(s)} \le \Ebb f^{(1 / 2)}$ by monotonicity, so $\Pbb[f^{(s)} = 1]
  \le \half$ $\implies$ $\Pbb[h^{(s)} = 1] \le \half + \eps \le \frac{3}{4}$ (as
  long as $\eps \le \quarter$) $\implies$ $\Pbb[h^{(s)} = 0] \ge \quarter$.
  Therefore $\Pbb[f^{(s)} = 1 \mid h^{(s)} = 0] \le 4\eps$.
  Therefore, there exists $u$ such that $h_{\onu} \equiv 0$ and
  $\Pbb[f_{\onu}^{(s)} = 1] \le 4\eps$.
  % TODO: check about the comments at beginning of lecture 14.
  By monotonicity, $\Pbb[f_{\onu}^{(p)} = 1] = \Ebb f_{\onu}^{(p)} \le 4\eps$,
  so choosing $\eps < \frac{\alpha}{5}$, we have that
  \[
    |\Ebb f_{\onu}^{(p)} - \Ebb f^{(p)}|
    > \eps
  ,\]
  contradicting \quasi{\eps, \rho, r}, where $r = r(\zeta, \eps)$.
\end{proof}

\begin{notation*}
  Let $\mathcal{B}$ be a family of subsets of $[n]$.
  Write $\ol{\mathcal{B}} = \{A \subset [n] : \exists B \in \mathcal{B}, B
  \subset A\}$, the \emph{upward closure} of $\mathcal{B}$.
\end{notation*}

\begin{fcthm}[Dinur, Friedgut]
  % Theorem 7.5
  \label{thm:7.5}
  For every $p \in \left( 0, \half \right)$ and $\eps > 0$, there exists $T$ such
  that for every intersecting family $\mathcal{A}$ of subsets of $[n]$ there
  exists $J \subset [n]$ of size at most $T$ and an intersecting family
  $\mathcal{B}$ of subsets of $J$ such that
  \[
    \mup(\mathcal{A} \setminus \ol{\mathcal{B}})
    \le \eps
  .\]
\end{fcthm}

\begin{proof}
  It is convenient to reformulate the statement in terms of Boolean functions.
  So call $f : \{0, 1\}^n \to \{0, 1\}$ \emph{intersecting} if for all $x, y$,
  $f(x) = f(y) = 1 \implies \exists i, x_i = y_i = 1$.
  
  Let $f : \{0, 1\}^n \to \{0, 1\}$ be a Boolean function and apply the
  regularity lemma with parameters $\left( \frac{\eps}{10}, p, r, \frac{\eps}{2}
  \right)$, where $r$ is to be chosen.
  That gives us $J$ of size at most $T(\eps, p, r)$ such that if $u$ is chosen
  $\mup$-randomly from $\{0, 1\}^J$, then
  \[
    \Pbb\left[\text{$f_{\onu}^{(p)}$ is not
    \glsref[quasi]{$\left(\frac{\eps}{10}, p, r\right)$-quasirandom}}\right]
    \le \frac{\eps}{2}
  .\]
  Define $g : \{0, 1\}^J \to \{0, 1\}$ by setting $g(u) = 1$ if $f_{\onu}^{(p)}$
  is \quasi{\frac{\eps}{10}, p, r}, and $\Ebb f_{\onu}^{(p)} \ge \frac{\eps}{2}$
  and $0$ otherwise.
  Then
  \[
    \Pbb[f^{(p)}(x) = 1 \text{ and } g(x|_J) = 0]
    \le \frac{\eps}{2} + \frac{\eps}{2}
  .\]
  (This corresponds to the statement that $|\mathcal{A} \setminus
  \ol{\mathcal{B}}| \le \eps$. $\mathcal{A} = \{A : f(\indicator{A}) = 1\}$,
  $\mathcal{B} = \{B : g(\indicator{B}) = 1\}$.)

  Note that $\ol{\mathcal{A}}$ is an intersecting family and $|\ol{\mathcal{A}}
  \setminus \ol{\mathcal{B}}| \ge |\mathcal{A} \setminus \mathcal{B}|$, so we
  may assume that $f$ is monotone, and hence that each $f_{\onu}$ is monotone.
  It remains to prove that $g$ is intersecting.
  Let $u, v \in \{0, 1\}^J$ such that $g(u) = g(v) = 1$.
  Then $f_{\onu}$ and $f_{\onu[v]}$ are \quasi{\frac{\eps}{10}, p, r} and $\Ebb
  f_{\onu}^{(p)}, \Ebb f_{\onu[v]}^{(p)} \ge \frac{\eps}{2}$.
  By \cref{lemma:7.4} for appropriate $r$, it follows that $\Ebb f_{\onu}^{(1 /
  2)}, \Ebb f_{\onu[v]}^{(1 / 2)} > \half$.

  By averaging, we can find $y, z \in \{0, 1\}^{[n] \setminus J}$ such that $y_i
  = 1 \iff z_i = 0$ and $f_{\onu}(y) = f_{\onu[v]}(z)) = 1$.
  Since $f$ is intersecting, there must exist $j$ such that $u_i = v_j = 1$, so
  $g$ is intersecting.
\end{proof}

\begin{fclemma}[The LYM inequality]
  % Lemma 7.6
  \label{lemma:7.6}
  Let $1 \le r < s \le n$ and let $\mathcal{A} \subset [n]^{(r)}$.
  Write $\partial_s \mathcal{A}$ for $\{B \in [n]^{(s)} : \exists A \in
  \mathcal{A}, A \subset B\}$.
  Then $\frac{|\partial_s \mathcal{A}|}{{n \choose s}} \ge
  \frac{|\mathcal{A}|}{{n \choose r}}$.
\end{fclemma}

\begin{proof}
  Let $\alpha = \frac{|\mathcal{A}|}{{n \choose r}}$ and $\beta =
  \frac{|\partial_s \mathcal{A}|}{{n \choose s}}$.
  Define a bipartite graph by joining $A \in [n]^{(r)}$ to $B \in [n]^{(s)}$ if
  and only if $A \subset B$.
  Now pick a random edge.
  The probability that it joins $\mathcal{A}$ to $\partial_s \mathcal{A}$ is
  exactly $\alpha$.
  It is also at most $\beta$.
  So $\alpha = \Pbb[\text{joins $\mathcal{A}$ to $\partial_s \mathcal{A}$}] \le
  \beta$.
\end{proof}