I'm trying to prove that $\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=\mu_{P}(p_1,p_2)\mu_{Q}(q_1,q_2)$, and my attempt is by induction. By definition, I get that:
$\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=-\sum_{(p_1,q_1)\leq(x,y)<(p_2,q_2)}\mu_{P\times Q}((p_1,q_1),(x,y))$, but for each $(x,y)$ there's a subposet of points between $(p_1,q_1)$ and $(x,y)$ so I use induction to assume the theorem works on each term and I get.
$\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=\mu_{P}(p_1,p_2)\mu_{Q}(q_1,q_2)=-\sum_{(p_1,q_1)\leq(x,y)<(p_2,q_2)}\mu_P(p_1,x)\mu_Q(q_1,y)$, now I can split the sums in $P$ and $Q$, getting:
$\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=-\sum_{p_1\leq x<p_2}\mu_P(p_1,x)\sum_{q_1\leq y<q_2}\mu_Q(q_1,y)$, but this gives me the product off by a minus sign:
$\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=-\mu_{P}(p_1,p_2)\mu_{Q}(q_1,q_2)$, and I can't see my mistake.
Edit: Forgot to mention that both $P$ and $Q$ are locally finite (let them be finite themselves even) so the subposets are finite.