Conditional Log-Density Convergence for the Poisson Reference

Let $A\subset F_{+}$ be finite. By [F1], $D(P_{A\cup C}\mathrm{\parallel }{Q}_{A\cup C})$ is finite, so the joint density $h_{A,C}$ relative to $Q_A\otimes Q_C$ exists. The marginal density $h_A$ is its $Q_C$ integral. On $0<h_A<\mathrm{\infty }$, [F2] gives the conditional density $k_A=h_{A,C}\mathrm{/}{h}_A$ relative to $Q_C$; the exceptional fibers have $P_A$-measure zero. The chain rule in [F2] gives $g(A)=E_{P}D(P_{C\mathrm{\mid }A}\mathrm{\parallel }{Q}_C)=D(P_{A\cup C}\mathrm{\parallel }{Q}_{A\cup C})-D(P_A\mathrm{\parallel }{Q}_A)$, so $g(A)$ is finite and nonnegative.

If $A\subseteq B\subset F_{+}$ are finite, apply the conditional-increment monotonicity in [F2] to observation of the cell configuration on $A$ and then on $B$. It yields $g(A)\le g(B)$.

To identify the supremum, let ${\mathrm{\Lambda }}_N=[0,N{)}^d$, the union of $N^d$ unit cells, and order those cells lexicographically. Repeated use of the conditional chain rule decomposes $D(P_{{\mathrm{\Lambda }}_N}\mathrm{\parallel }{Q}_{{\mathrm{\Lambda }}_N})$ as the sum of $g(B_i)$ over its cells $i$, where $B_i$ is the finite set of earlier cells in ${\mathrm{\Lambda }}_N$. Each $B_i$ is a finite subset of $F_{+}$ after translating cell $i$ to $C$, so $g(B_i)\le {\sup }_{A}g(A)$ by stationarity and step 2.1. Hence $D(P_{{\mathrm{\Lambda }}_N}\mathrm{\parallel }{Q}_{{\mathrm{\Lambda }}_N})\mathrm{/}{N}^d\le {\sup }_{A}g(A)$. Conversely, fix finite $A\subset F_{+}$. For every cell $i$ in the inner cube whose distance from the boundary exceeds the largest coordinate magnitude in $A$, the translated set $i+A$ lies in $B_i$. By step 2.1 and stationarity, $g(B_i)\ge g(A)$. The proportion of these interior cells tends to one, and all other increments are nonnegative by step 1.1. Taking $N\to \mathrm{\infty }$ and applying [F1] gives $\gamma \ge g(A)$. Taking the supremum over $A$ proves ${\sup }_{A}g(A)=\gamma$.

Choose an increasing sequence of finite pasts $A_m$ exhausting $F_{+}$ and set $R=P_{F_{+}}\otimes Q_C$. Lift each $k_{A_m}$ to the common space $({\mathrm{\Omega }}_{F_{+}}\times E_C,R)$. Consistency of the conditional densities shows $E_R[k_{A_{m+1}}\mid {\mathcal{G}}_m]=k_{A_m}$, where ${\mathcal{G}}_m$ is generated by $A_m$ and $C$; also $E_R{k}_{A_m}=1$. By step 3.1, $E_R[k_{A_m}\log k_{A_m}]=g(A_m)\le \gamma$. For $K>1$, splitting at $k_{A_m}=1$ and using $-u\log u\le 1\mathrm{/}e$ gives ${\int }_{\{k_{A_m}>K\}}{k}_{A_m}dR\le \frac{\gamma +1\mathrm{/}e}{\log K}\mathrm{.}$ Thus these martingale densities are uniformly integrable. By [F3] they converge almost surely and in $L^1(R)$ to $k_{\mathrm{\infty }}\ge 0$ with $E_R{k}_{\mathrm{\infty }}=1$.

For any finite-past cylinder event and any event in $C$, integrate $k_{\mathrm{\infty }}$ over their product. $L^1(R)$ convergence makes this equal the limit of the same integral against $k_{A_m}$, which is the corresponding joint probability under $P$. These rectangles form a generating pi-system, so [F6] identifies $k_{\mathrm{\infty }}R$ with $P_{F_{+}\cup C}$. Therefore $k_A=E_R[k_{\mathrm{\infty }}\mid {\mathcal{G}}_A]$ for every finite $A\subset F_{+}$. Fatou applied to $u\log u+1\mathrm{/}e\ge 0$ gives $D(P_{F_{+}\cup C}\mathrm{\parallel }{P}_{F_{+}}\otimes Q_C)\le {\mathrm{lim\hspace{0.17em}inf}}_{m}g(A_m)=\gamma$. Conversely, its variational supremum contains all finite-past tests, so step 3.1 gives the reverse inequality. Hence $E_P\log k_{\mathrm{\infty }}=\gamma$.

The $L^1(R)$ convergence of $k_{A_m}$ implies convergence in $R$-probability. Since the full joint law has density $k_{\mathrm{\infty }}$ relative to $R$, absolute continuity transfers this to convergence in $P$-probability. Also $k_{\mathrm{\infty }}>0$ $P$-almost surely, so continuity of $\log$ on $(0,\mathrm{\infty })$ gives $\log k_{A_m}\to \log k_{\mathrm{\infty }}$ in $P$-probability. For $t\ge 1$, $E_P\text{\hspace{0.17em}⁣}[(-\log k_A{)}^{+}{\mathbf{1}}_{\{(-\log k_A{)}^{+}>t\}}]={\int }_{\{0<k_A<e^{-t}\}}{k}_A(-\log k_A)dR\le t{e}^{-t}\mathrm{.}$ Thus the negative parts are uniformly integrable; [F4] gives their $L^1(P)$ convergence.

Since $E_P\log k_{A_m}=g(A_m)\to \gamma =E_P\log k_{\mathrm{\infty }}$ and the negative parts converge in $L^1$, the expectations of the positive parts converge as well. These positive parts converge in probability. For nonnegative variables $Y_m\to Y$ in probability with $E{Y}_m\to EY<\mathrm{\infty }$, bounded truncations satisfy $E(Y_m\wedge K)\to E(Y\wedge K)$: for any $\eta >0$ the difference is at most $\eta +KP(\mathrm{\mid }{Y}_m-Y\mathrm{\mid }>\eta )$. Consequently $E(Y_m-K{)}^{+}\to E(Y-K{)}^{+}$; choosing $K$ large and then bounding $Y_m{\mathbf{1}}_{Y_m>2K}\le 2(Y_m-K{)}^{+}$ proves uniform integrability, including the finite initial segment. Applying [F4] gives $L^1$ convergence of the positive parts. Together with the negative parts this proves $\log k_{A_m}\to \log k_{\mathrm{\infty }}$ in $L^1(P)$.

For any finite $B\supseteq A_m$, the tower property gives $k_B=E_R[k_{\mathrm{\infty }}\mid {\mathcal{G}}_B]$ and $k_{A_m}=E_R[k_{\mathrm{\infty }}\mid {\mathcal{G}}_m]$. Conditional expectation is an $L^1$ contraction, so ${\sup }_{B\supseteq A_m}\mathrm{\parallel }{k}_B-k_{\mathrm{\infty }}{\mathrm{\parallel }}_{L^1(R)}\le 2\mathrm{\parallel }{k}_{A_m}-k_{\mathrm{\infty }}{\mathrm{\parallel }}_{L^1(R)}⟶0.$ Absolute continuity of $P=k_{\mathrm{\infty }}R$ transfers this uniform probability convergence to $P$. The negative-part tail bound in step 6.1 is uniform over all finite $B$, while $g(A_m)\le g(B)\le \gamma$ by steps 2.1–3.1. The truncation argument in step 7.1 therefore applies to every sequence $B_m\supseteq A_m$ with $m\to \mathrm{\infty }$; if the asserted directed $L^1$ convergence were not uniform, choosing a contradicting sequence $B_m$ would violate that argument. This proves the directed form.