Local topology, entropy compactness and path-space estimates

The result is immediate when $T=0$. Suppose $T>0$, fix $n\ge 0$, and let $t_k=kT\mathrm{/}{2}^n$ for $0\le k\le 2^n$. Put $S_n={\max }_{0\le k\le 2^n}\mathrm{\mid }{M}_{t_k}\mathrm{\mid }$, which is square-integrable because it is bounded by the finite sum ${\sum }_k\mathrm{\mid }{M}_{t_k}\mathrm{\mid }$. For $r>0$, let $\tau$ be the first grid index $k$ with $\mathrm{\mid }{M}_{t_k}\mathrm{\mid }\ge r$, or leave it undefined if no grid value reaches $r$. Each event $C_k=\{\tau =k\}$ belongs to ${\mathcal{F}}_{t_k}$. By [F1] and [F2], $\mathrm{\mid }{M}_{t_k}\mathrm{\mid }=\mathrm{\mid }\mathbb{E}[M_T\mid {\mathcal{F}}_{t_k}]\mathrm{\mid }\le \mathbb{E}[\mathrm{\mid }{M}_T\mathrm{\mid }\mid {\mathcal{F}}_{t_k}]$ almost surely. Thus [F7] gives $\mathbb{E}[\mathrm{\mid }{M}_T\mathrm{\mid }{\mathbf{1}}_{C_k}]=\mathbb{E}[\mathbb{E}[\mathrm{\mid }{M}_T\mathrm{\mid }\mid {\mathcal{F}}_{t_k}]{\mathbf{1}}_{C_k}]\ge \mathbb{E}[\mathrm{\mid }{M}_{t_k}\mathrm{\mid }{\mathbf{1}}_{C_k}]\ge r\mathbb{P}(C_k)$. Summing over the finite grid yields $r\mathbb{P}(S_n\ge r)\le \mathbb{E}[\mathrm{\mid }{M}_T\mathrm{\mid }{\mathbf{1}}_{\{S_n\ge r\}}]\mathrm{.}$

Apply [F4] to $S_n^2$, use $\mathbb{P}(S_n>r)\le \mathbb{P}(S_n\ge r)$ and integrate the bound from step 1.1. [F5] for the nonnegative measurable function $(r,\omega )\mapsto \mathrm{\mid }{M}_T(\omega )\mathrm{\mid }{\mathbf{1}}_{\{S_n(\omega )\ge r\}}$ gives $\mathbb{E}{S}_n^2={\int }_0^{\mathrm{\infty }}2r\mathbb{P}(S_n>r)dr\le 2\mathbb{E}[\mathrm{\mid }{M}_T\mathrm{\mid }{S}_n]\mathrm{.}$ By [F3], the right side is at most $2(\mathbb{E}\mathrm{\mid }{M}_T{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}(\mathbb{E}{S}_n^2{)}^{1\mathrm{/}2}$. If $\mathbb{E}{S}_n^2=0$ the desired finite-grid bound is immediate; otherwise divide by its square root and square to obtain $\mathbb{E}{S}_n^2\le 4\mathbb{E}\mathrm{\mid }{M}_T{\mathrm{\mid }}^2$.

The dyadic grids are nested, so $S_n\uparrow S_{\ast }:={\sup }_{t\in D_T}\mathrm{\mid }{M}_t\mathrm{\mid }=S_T$. For every $\omega \in {\mathrm{\Omega }}_0$ and each $t<T$, there are dyadic grid times decreasing to $t$, and right-continuity makes their process values converge to $M_t(\omega )$; time $T$ is in every grid. Hence $S_{\ast }$ equals the pathwise supremum on $[0,T]$ almost surely. Apply [F6] to $S_n^2$ and take limits in the finite-grid estimate from step 2.1 to obtain the stated inequality.

Write $p_x=P_X(x)$ and $q_x=Q_X(x)$. Strict positivity of $Q$ gives $q_x>0$ and $Q_{Y\mathrm{\mid }x}(y)>0$ for each $x,y$. If $P(x,y)>0$, then $p_x>0$ and $\ln \frac{P(x,y)}{Q(x,y)}=\ln \frac{p_x}{q_x}+\ln \frac{P_{Y\mathrm{\mid }x}(y)}{Q_{Y\mathrm{\mid }x}(y)}$ by [F2]. Multiply by $P(x,y)$ and sum over positive joint atoms. For each $x$ with $p_x>0$, summing the first term over $y$ gives $p_x\ln (p_x\mathrm{/}{q}_x)$, and summing the second gives $p_{x}D(P_{Y\mathrm{\mid }x}\mathrm{\parallel }{Q}_{Y\mathrm{\mid }x})$. Atoms with $p_x=0$ contribute no terms. This is exactly the claimed chain rule.

If $C$ is empty or all of $I\times J$, then $P(C)=Q(C)$ is respectively $0$ or $1$, so the binary divergence is $0$ under the stated zero-weight convention. The grouping bound is then $D(P\mathrm{\parallel }Q)\ge 0$, with equality exactly when $P=Q$ by [F1]. For a nonempty proper $C$, strict positivity gives $0<Q(C)<1$. Put $a=P(C)$ and $b=Q(C)$. When $a>0$, the conditional laws on $C$ are $P_C(z)=P(z)\mathrm{/}a$ and $Q_C(z)=Q(z)\mathrm{/}b$; the latter is strictly positive on $C$. The logarithmic identity $\ln (P(z)\mathrm{/}Q(z))=\ln (a\mathrm{/}b)+\ln (P_C(z)\mathrm{/}{Q}_C(z))$ for each $z\in C$ with $P(z)>0$ yields ${\sum }_{z\in C:P(z)>0}P(z)\ln \frac{P(z)}{Q(z)}=a\ln \frac{a}{b}+aD(P_C\mathrm{\parallel }{Q}_C)\mathrm{.}$ If $a=0$, both the sum and the weighted conditional term are zero, so the same identity holds with that term omitted. Applying the identical calculation to $C^c$ and adding gives $D(P\mathrm{\parallel }Q)=d(a\mathrm{\parallel }b)+aD(P_C\mathrm{\parallel }{Q}_C)+(1-a)D(P_{C^c}\mathrm{\parallel }{Q}_{C^c})$, with zero-weight conditional terms omitted. Both weighted conditional divergences are nonnegative by [F1], which proves the grouping inequality, including $a=0$ or $1$. Equality holds precisely when every conditional divergence with positive weight vanishes; by the equality clause of [F1], this is equivalent to agreement of the conditional probability vectors on each positive-$P$-mass block. Zero-$P$-mass blocks impose no conditional-law condition.

If $\sigma \mathrm{\ne }\eta$, some coordinate differs, so $\rho (\sigma ,\eta )>0$; symmetry is immediate, and the coordinatewise triangle inequality gives the triangle inequality after summing. If two configurations agree on $x_1,\dots ,x_N$, then $\rho \le {\sum }_{k>N}{2}^{-k}=2^{-N}$. Conversely, if $\rho (\sigma ,\eta )<2^{-N}$, no coordinate among $x_1,\dots ,x_N$ differs, because a mismatch at $x_k$ contributes at least $2^{-N}$. These bounds show that metric balls and finite-coordinate basic neighborhoods generate the same topology.

If ${\mu }_n{\to }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mu$, every cylinder indicator is bounded and continuous by [F2], so its integrals converge. Every bounded function depending on finitely many coordinates is a finite linear combination of cylinder indicators, and its integrals therefore converge as well.

Suppose an open cover of $\mathrm{\Omega }$ has no finite subcover. Let $C_0=\mathrm{\Omega }$. Recursively, if $C_{N-1}$ is the cylinder fixing the first $N-1$ spins and has no finite subcover, its finitely many children obtained by assigning ${\sigma }_{x_N}=s$ cannot all have finite subcovers; otherwise their finite union would cover $C_{N-1}$. Choose the least $s$ in the fixed order whose child still has no finite subcover and call that child $C_N$. These nested cylinders determine a configuration $\sigma$. A cover member containing $\sigma$ contains a basic neighborhood fixing finitely many sites; for sufficiently large $N$, $C_N$ lies inside that neighborhood and hence has a one-member subcover, a contradiction. Thus every open cover has a finite subcover and $\mathrm{\Omega }$ is compact.

Each cylinder is clopen by [F2], and cylinders form a countable basis: ${\mathbb{Z}}^d$ has countably many finite subsets, and each finite coordinate set has finitely many spin assignments. Therefore every open set is a countable union of cylinders and belongs to the cylinder sigma-algebra, while every cylinder is Borel. Hence the two sigma-algebras are equal.

Fix a bounded continuous $g:\mathrm{\Omega }\to \mathbb{R}$ and $\epsilon >0$. For each $\sigma$, continuity and the cylinder basis give a cylinder neighborhood $C_{\sigma }$ of $\sigma$ on which $\mathrm{\mid }g(\eta )-g(\sigma )\mathrm{\mid }<\epsilon \mathrm{/}3$. Compactness from step 2.1 gives a finite subcover $C_1,\dots ,C_r$. Let $F$ be the union of the finitely many coordinate sets fixed by these cylinders. If $\eta ,\zeta$ agree on $F$, choose a $C_i$ containing $\eta$; because $C_i$ depends only on a subset of $F$, it also contains $\zeta$, so $\mathrm{\mid }g(\eta )-g(\zeta )\mathrm{\mid }<2\epsilon \mathrm{/}3$. Choose $N$ with $F\subseteq \{x_1,\dots ,x_N\}$ and fix one default spin $s_0\in S$. Replace all coordinates after $x_N$ by $s_0$ to define a canonical extension $\widehat{\sigma }$ of each first-$N$ prefix, and set $g_N(\sigma )=g(\widehat{\sigma })$. Then $g_N$ depends only on the first $N$ coordinates and ${\sup }_{\sigma }\mathrm{\mid }g(\sigma )-g_N(\sigma )\mathrm{\mid }<\epsilon$.

For every $n$, $\mathrm{\mid }\int gd{\mu }_n-\int g_{N}d{\mu }_n\mathrm{\mid }\le \epsilon$ and the same bound holds for $\mu$. Step 1.2 gives $\int g_{N}d{\mu }_n\to \int g_{N}d\mu$, so the limit superior of $\mathrm{\mid }\int gd{\mu }_n-\int gd\mu \mathrm{\mid }$ is at most $2\epsilon$. Since $\epsilon >0$ is arbitrary, weak convergence follows. Conversely, if ${\mu }_n$ converges weakly, indicators of all cylinders are bounded continuous by [F2], so their probabilities converge; finite sums give every finite marginal, hence local weak convergence by [F1].

Put ${\lambda }_i={\sum }_{j\mathrm{\ne }i}{L}_{ij}$ and $q={\max }_i{\lambda }_i$. If $q=0$, all off-diagonal entries vanish and the zero row sums force $L=0$. On the finite probability space $(E,2^E,\nu )$ put $Y_0(\omega )=\omega$ and $X_t=Y_0$ for all $t$. Then $N_t=0$, paths are constant, the conditional transition is the identity, and $e^{tL}=I$, giving the conclusion in this case.

Suppose $q>0$. For $i\mathrm{\ne }j$, $R_{ij}=L_{ij}\mathrm{/}q\ge 0$, while $R_{ii}=1-{\lambda }_i\mathrm{/}q\ge 0$; each row sums to $1$. Let ${\mu }_q(dw)=q{e}^{-qw}dw$ on $(0,\mathrm{\infty })$, a probability measure because its density integrates to $1$. On $E\times (0,\mathrm{\infty })$ use initial law $\nu \otimes {\mu }_q$ and transition kernel $K((i,w),\{j\}\times B)=R_{ij}{\mu }_q(B)$ for Borel $B\subseteq (0,\mathrm{\infty })$. For any Borel subset of the product, its kernel value is the finite sum over $j$ of $R_{ij}$ times the ${\mu }_q$-measure of the corresponding section; thus it is measurable, nonnegative and has total mass $1$.

Apply [F2] to this initial law and kernel, and write the resulting coordinates as $Z_n=(Y_n,W_{n+1})$. Their finite-prefix law factors as the Markov-chain prefix law for $Y$ times ${\prod }_{k=1}^{n+1}{\mu }_q(d{W}_k)$, because every transition kernel chooses the next state with probabilities $R_{ij}$ and the next waiting time with the same factor ${\mu }_q$, independently of the current waiting time. Cylinder events generate the two coordinate sigma-algebras, so [F3] extends this factorization to independence of the full sequence $Y$ from the iid family $(W_k)$. The first coordinate is an $R$-Markov sequence with initial law $\nu$, as required by [F1].

Each $W_k$ has probability $e^{-1}$ of being at least $q^{-1}$. For any fixed $r$, independence gives $\mathbb{P}(W_k<q^{-1}\text{ for }r\le k\le n)=(1-e^{-1}{)}^{n-r+1}\to 0$. Continuity of probability for decreasing events, obtained by applying [F4] to complements, shows that the probability of no such event after $r$ is zero. A countable union over $r$ shows that $W_k\ge q^{-1}$ occurs infinitely often almost surely, and therefore ${\sum }_k{W}_k=\mathrm{\infty }$ almost surely. On every bounded time interval there are consequently finitely many event times; the convention $X_t=Y_{N_t}$ makes paths right-continuous with left limits.

For $n\ge 1$, convolution of the exponential densities gives the arrival-time density $f_{T_n}(u)=\frac{q^n{u}^{n-1}{e}^{-qu}}{(n-1)!},u>0.$ Indeed the formula holds for $n=1$, and convolving it with $q{e}^{-q(t-u)}$ gives $q^{n+1}{e}^{-qt}{\int }_0^t{u}^{n-1}du\mathrm{/}(n-1)!=q^{n+1}{t}^n{e}^{-qt}\mathrm{/}n!$. Since $W_{n+1}$ is independent of $T_n$, for $n\ge 1$, $\mathbb{P}(N_t=n)={\int }_0^t{f}_{T_n}(u)\mathbb{P}(W_{n+1}>t-u)du=e^{-qt}\frac{(qt{)}^n}{n!};$ for $n=0$ the probability is $\mathbb{P}(W_1>t)=e^{-qt}$.

The event process has independent rate-$q$ Poisson increments. To see the conditional assertion needed here, on $\{N_s=n\}$ the event-clock history gives $T_n\le s<T_{n+1}$; by the exponential memoryless identity $\mathbb{P}(W_{n+1}-(s-T_n)>u\mid W_{n+1}>s-T_n)=e^{-qu}$, the residual time after $s$ is rate-$q$ exponential independent of that history. The later waiting times are independent rate-$q$ exponentials as well. Repeating the calculation in step 3.2 shows that, conditional on the clock history up to $s$, $N_{s+h}-N_s$ has probabilities $e^{-qh}(qh{)}^k\mathrm{/}k!$, independent of that history. The clock is independent of $Y$ by step 2.1 and the uniformized convention [F1].

Let $H_s=(Y_0,\dots ,Y_{N_s})$ be the finite embedded-state prefix and put ${\mathcal{G}}_s=\sigma (N_u:0\le u\le s,H_s)$. Since $N_u\le N_s$ for $u\le s$, the construction filtration ${\mathcal{F}}_s=\sigma (N_u,X_u:0\le u\le s)$ is contained in ${\mathcal{G}}_s$. Given ${\mathcal{G}}_s$ and $X_s=i$, $k$ further clock events have probability $e^{-qh}(qh{)}^k\mathrm{/}k!$ by step 4.1, and the embedded chain has $k$-step transition matrix $R^k$. Hence $\mathbb{P}(X_{s+h}=j\mid {\mathcal{G}}_s)={\sum }_{k=0}^{\mathrm{\infty }}{e}^{-qh}\frac{(qh{)}^k}{k!}(R^k{)}_{ij}=(e^{-qh}{e}^{qhR}{)}_{ij}=(e^{hL}{)}_{ij},$ since $q(R-I)=L$. This conditional law depends on ${\mathcal{G}}_s$ only through $X_s$; [F5] applied to the integrable indicator ${\mathbf{1}}_{\{X_{s+h}=j\}}$ and ${\mathcal{F}}_s\subseteq {\mathcal{G}}_s$ gives the required identity for all $s\le t$. The matrix series converges absolutely because $E$ is finite.

The constructed $X$ is adapted to ${\mathcal{F}}_t$ by definition. Step 3.1 gives nonexplosion and càdlàg paths; step 5.1 gives its all-pairs conditional transition matrices. Thus it is a uniformized realization for every finite rate matrix, including reducible matrices and states with ${\lambda }_i=0$.

Fix $t>0$ and a bounded function $a:E\to \mathbb{R}$. On $[0,t]$, approximate $a(X_u)$ by the process with value $a(X_{kt\mathrm{/}{2}^n})$ on $((k-1)t\mathrm{/}{2}^n,kt\mathrm{/}{2}^n]$, $1\le k\le 2^n$, and value $a(X_0)$ at zero. Each approximation is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable by adaptedness of $X$ in [F1], and converges pointwise by right continuity. The construction uses constant paths on its exceptional explosion event, so this argument applies there too. Its time integral is the ${\mathcal{F}}_t$-measurable sum $(t\mathrm{/}{2}^n){\sum }_{k=1}^{2^n}a(X_{kt\mathrm{/}{2}^n})$. These sums converge to ${\int }_0^{t}a(X_u)du$: on a path with $J$ jumps, the error is at most $2\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}Jt\mathrm{/}{2}^n$, since only intervals containing a jump contribute. Thus the time integral is ${\mathcal{F}}_t$-measurable, without completing the filtration. Taking $a=g$ proves that $M$ is adapted, also at $t=0$. By [F1], $\mathbb{E}[g(X_{s+r})\mid {\mathcal{F}}_s]=(e^{rL}g)(X_s)$. The joint measurability just proved and boundedness allow [F5] on finite time intervals. Testing against each event in ${\mathcal{F}}_s$ gives $\mathbb{E}\text{\hspace{0.17em}⁣}[{\int }_s^{s+h}g(X_u)du|{\mathcal{F}}_s]={\int }_0^h(e^{rL}Lf)(X_s)dr=(e^{hL}f-f)(X_s),$ where the last equality follows by termwise differentiation of the absolutely convergent finite-matrix exponential series. Also $\mathbb{E}[f(X_{s+h})\mid {\mathcal{F}}_s]=(e^{hL}f)(X_s)$. Subtracting and using adaptedness proves $\mathbb{E}[M_{s+h}\mid {\mathcal{F}}_s]=M_s$ for every $s,h$. On $[0,T]$, $\mathrm{\mid }{M}_t\mathrm{\mid }\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+T\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}$, so every $M_t$ is square-integrable and [F2] gives the all-pairs martingale assertion.