Proof of Uniqueness for the Forward Equation of a Bounded Jump-Rate Family on a Finite Set

Throughout, integrals over compact intervals are the Lebesgue integrals over the compact interval in question, and we use linearity and monotonicity of the integral freely for bounded measurable integrands on such intervals (such an integrand is integrable, the interval having finite measure by claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval), including the bound $\bigl|\int f\bigr|\le\int|f|$ of claim 2 of that theorem.

Step 1: the gap function. For $t\in[r,T]$ put

By condition (i) for $\mu$ and $\nu$ and by Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions (differences, absolute values, and finite sums of bounded measurable functions), $d$ is bounded and measurable on $[r,T]$ with respect to $\mathcal{B}_{[r,T]}$, and $d\ge0$.

Step 2: the integral inequality. Fix $y\in E$ and let $\mathbf{1}_y:E\to\mathbb{R}$ be the function equal to $1$ at $y$ and $0$ elsewhere. From the definition of $\mathcal{L}_u$,

so that $|\mathcal{L}_u\mathbf{1}_y(x)|\le\Lambda$ for all $u\in[0,T]$ and $x\in E$, by the rate bound in the hypotheses (each $q_u(x,y)$ is nonnegative and at most the sum bounded by $\Lambda$). Applying condition (ii) with $F=\mathbf{1}_y$ to $\mu$ and to $\nu$, using $\mu_r=\nu_r$, and subtracting (linearity of the integral, both integrands being bounded and measurable), for every $t\in[r,T]$,

The integrand is bounded in absolute value by $\Lambda\,d(u)$, by Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, so $|\mu_t(y)-\nu_t(y)|\le\Lambda\int_{[r,t]}d(u)\,du$. Summing over the $|E|$ points $y\in E$ (with $|E|$ the number of elements of $E$) gives

where for $t=r$ both sides are $0$.

Step 3: Gronwall. Define $d^0:[0,T]\to\mathbb{R}$ by $d^0(t)=0$ for $t\in[0,r)$ and $d^0(t)=d(t)$ for $t\in[r,T]$. Then $d^0$ is bounded and nonnegative, and it is measurable with respect to $\mathcal{B}_{[0,T]}$: for a real number $a\ge0$ the set $\{t\in[0,T]:d^0(t)>a\}=\{t\in[r,T]:d(t)>a\}$ belongs to $\mathcal{B}_{[r,T]}$, hence is a Borel set (claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval) contained in $[0,T]$, hence belongs to $\mathcal{B}_{[0,T]}$; for $a<0$ the set is all of $[0,T]$; so $d^0$ is measurable by the generator criterion of Measurable Function and Real-Valued Measurable Function. For $t\in(r,T]$, the zero extensions to $\mathbb{R}$ of the restriction of $d^0$ to $[0,t]$ and of the restriction of $d$ to $[r,t]$ are the same function on $\mathbb{R}$ (both vanish off $[r,t]$ and agree with $d$ on $[r,t]$), so by claim 2 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, applied on $[0,t]$ and on $[r,t]$,

Combining with Step 2, $d^0(t)\le|E|\Lambda\int_{[0,t]}d^0(u)\,du$ for $t\in(r,T]$, and also for $t=r$, where $d^0(r)=d(r)=0$ because $\mu_r=\nu_r$ while the right side is nonnegative; for $t\in[0,r)$ the same inequality holds because its left side is $0$ and its right side is the integral of a nonnegative function (monotonicity), or $0$ when $t=0$. Thus the hypotheses of Gronwall's lemma for bounded measurable functions hold on $[0,T]$ with the function there taken to be $d^0$, its additive constant taken to be $0$, and its multiplicative constant $c=|E|\Lambda$, and it yields $d^0(t)\le0\cdot\exp(ct)=0$ for every $t\in[0,T]$. Since $d^0\ge0$, $d(t)=0$ for every $t\in[r,T]$, that is, $\mu_t(x)=\nu_t(x)$ for every $x\in E$ and every $t\in[r,T]$.