Let $[n] = \{1 , \dots, n\}$ and let $\mathcal{A} \in \mathcal{P}([n])$ be an increasing set, i.e. if $x \in \mathcal{A}$ and $x \subset y \subset [n]$ then $y \in \mathcal{A}$.
For $p \in [0,1]$, define the $p$-biased measure of a subset $A \subset [n]$ to be:
$$\mu_p(A) = p^{|A|}(1-p)^{n - |A|}$$
For a family of sets $\mathcal{A} \subset \mathcal{P}([n])$, define:
$$\mu_p(\mathcal{A}) = \sum_{A \in \mathcal{A}} \mu_p(A)$$
Theorem 7 says that for $0 < p <1$ and $\mathcal{A}, \mathcal{B} \subset \mathcal{P}([n])$ both increasing families of sets, then:
$$\mu_p(\mathcal{A})\mu_p(\mathcal{B}) \leq \mu_p(\mathcal{A} \cap \mathcal{B})$$
With some work, I've managed to show that:
$$\mu_p(\mathcal{A})\mu_p(\mathcal{B}) = \sum_{A \in \mathcal{A} \backslash \mathcal{B} \\ B \in\mathcal{B} \backslash \mathcal{A}}{\mu_p(\mathcal{A})\mu_p(\mathcal{B})} + \sum_{C \in \mathcal{A} \cap \mathcal{B}}{\mu_p(C) \sum_{A \in \mathcal{A} \backslash \mathcal{B} \\ B \in\mathcal{B} \backslash \mathcal{A}} (\mu_p(A) + \mu_p(B)}) + (\mu_p(\mathcal{A} \cap \mathcal{B})^2$$
However, at this point I am stuck and I'm not sure how to proceed, or how to use the fact that $\mathcal{A}, \mathcal{B}$ are increasing sets. How can I use this piece of information?