A proper Lyapunov bound prevents reaction-network explosion

Fix $n\ge \max \{1,V(x_0)\}$ and set $K_n=\{x\in E:V(x)\le n\}$. This is finite by properness. The set $D_n=\{x+{\nu }_j:x\in K_n,a_j(x)>0,1\le j\le R\}\setminus K_n$ of one-reaction exit states is finite, and $F_n=K_n\cup D_n$ is finite. The rate bound $M_n:={\max }_{x\in K_n}\lambda (x)$ is therefore finite. For fixed ordered channel labels $j_1,\dots ,j_m$, the density of event times $0<t_1<\cdots <t_m\le T$ before the first exit is the product of the successive channel rates and exponential no-event factors. Summing over channel labels bounds this density by $M_n^m$, since at each event the sum of active channel rates is at most $M_n$. The ordered simplex has volume $T^m\mathrm{/}m!$, so the probability of at least $m$ events before exit by $T$ is at most $(M_{n}T{)}^m\mathrm{/}m!$, which tends to zero. Thus infinitely many jumps cannot occur in finite time while the process remains in $K_n$.

Stop the process at ${\tau }_n:=\inf \{t\ge 0:X_t\mathrm{\notin }{K}_n\}$. The stopped path takes values in the finite set $F_n$ and is frozen after exit. Let $C_n={\max }_{y\in F_n}V(y)$. Conditional on the reaction history through time $s$, partition on the finite vector of last-used stream indices; on each part [F3] gives independence of all unused coordinates. If $s$ falls within a holding interval, the residual exponential wait has the same law because $e^{-\lambda (w+u)}\mathrm{/}{e}^{-\lambda w}=e^{-\lambda u}$. Thus for $x=X_s\in K_n$, the density of one channel-$j$ event at elapsed time $u$ and no earlier event is $a_j(x)e^{-\lambda (x)u}$. The probability of at least two events within $h$ is at most $(M_{n}h{)}^2\mathrm{/}2$ by integrating the summed first-two-event density, bounded by $M_n^2$, over the ordered two-time simplex. Since $V\le C_n$ on $F_n$, the first-event term differs from $h(QV)(x)$ by at most $C_n{M}_n^2{h}^2\mathrm{/}2$, using $\mathrm{\mid }1-e^{-z}\mathrm{\mid }\le z$ for $z\ge 0$; the multiple-event remainder is bounded by $2C_n(M_{n}h{)}^2\mathrm{/}2$. Thus, uniformly over $x\in K_n$, the conditional mean increment is $h{1}_{\{s<{\tau }_n\}}(QV)(X_s)+O(h^2)$. This conditional mean is the explicit kernel average from the independent stream and survival ratio just computed; integrating it over finite reaction histories gives the unconditional mean increment. Sum those means on a uniform partition of $[0,t]$ and let the mesh tend to zero to obtain $\mathbb{E}[V(X_{t\wedge {\tau }_n})]=V(x_0)+\mathbb{E}{\int }_0^{t\wedge {\tau }_n}(QV)(X_s)ds\mathrm{.}$ The error sum is $O(th)$; the pathwise Riemann sums converge by step 1.1 and are bounded by $t{\max }_{x\in K_n}\mathrm{\mid }QV(x)\mathrm{\mid }$, so [F5] passes their expectations to the limit. The same bounded Riemann-sum argument interchanges time integration and expectation for this bounded stopped integrand.

Set $g_n(t)=\mathbb{E}[V(X_{t\wedge {\tau }_n})]$. From step 2.1, $QV\le cV$, and $V\ge 1$, we obtain $g_n(t)\le V(x_0)+c{\int }_0^t{g}_n(s)ds\mathrm{.}$ Here $g_n$ is bounded on each finite interval because $X_{t\wedge {\tau }_n}\in F_n$. Iterating the integral inequality $m$ times bounds it by $V(x_0){\sum }_{k=0}^{m-1}(ct{)}^k\mathrm{/}k!+C_n(ct{)}^m\mathrm{/}m!$; the last term tends to zero and the sum tends to $V(x_0)e^{ct}$. Hence $g_n(t)\le V(x_0)e^{ct}$.

On $\{{\tau }_n\le T\}$ the exit state satisfies $V(X_{{\tau }_n})>n$, so $n\mathbb{P}({\tau }_n\le T)\le g_n(T)\le V(x_0)e^{cT}$.

Fix any finite $T>0$. If ${\tau }_{\mathrm{\infty }}\le T$ but ${\tau }_n>T$ for some integer $n\ge \max \{1,V(x_0)\}$, infinitely many events occur in finite time while the process remains in $K_n$, an event of probability zero by step 1.1. A countable union over such $n$ is still null, so outside it explosion by $T$ forces ${\tau }_n\le T$ for every sufficiently large integer $n$. Step 4.1 gives $\mathbb{P}({\tau }_{\mathrm{\infty }}\le T)\le V(x_0)e^{cT}\mathrm{/}n$ for arbitrarily large $n$, hence this probability is zero. Taking integer $T\ge 1$ proves $\mathbb{P}({\tau }_{\mathrm{\infty }}=\mathrm{\infty })=1$. The argument uses Countable Choice and Dependent Choice only as declared in [A1] and [A2]; it uses no Axiom of Choice.