Solution
= Solution
The same mixing argument with a fixed $q^0\in\mathcal Q$ shows that
$$
q^TZ\geq0
$$
for every $q\geq0$ in $\ker P^T$, including boundary <vectors>. By the <finite-state superhedging alternative>, there is $\phi\in\mathbb R^d$ such that
$$
\boxed{Z-P\phi\geq0,}
$$
which is the asserted almost-sure inequality. No strict residual is required here.