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A Markov jump process, considered up to its explosion time, is an adapted càdlàg piecewise-constant process whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. Describe its rates by an off-diagonal kernel $q(x,dy)$ with finite total rate $q(x,E)$. In state $x$ the holding time has the exponential distribution with this rate, followed by a destination distributed as $q(x,dy)/q(x,E)$; a zero-rate state is absorbing. Its Markov jump-process generator acts byThe Markov property must hold relative to the specified filtration, not merely the natural filtration: future-revealing information could change the jump compensator.

Let $\mu$ count jumps, marked by their destination states. For this jump measure of a Markov jump process, the predictable compensator isThe compensated random-measure theorem says that a predictable $H(s,y)$ with $\mathbb{E}{\int }_0^t\int |H|d\nu <\infty$ gives a true martingaleThe same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $1_A{1}_{(s,t]}H$, for $A\in {\mathcal{F}}_s$, shows that each increment has zero conditional expectation. Taking $H(s,y)=f(y)-f(X_{s-})$ gives the generator local martingale $f(X_t)-f(X_0)-{\int }_0^{t}Qf(X_s)ds$. This states the integrability conditions in the compensated jump-measure local martingale result explicitly.

For the birth-process claim first take the customary fixed initial state $X_0=i$. Write $N_t=X_t-i$, ${\Lambda }_t={\int }_0^t\lambda (X_s)ds$ and $a=e^{\theta }-1$. The compensator of $N$ is $\Lambda$, whose integrand may equivalently use $X_{s-}$ because jump times have zero Lebesgue measure. A birth multiplies $e^{\theta X}$ by $e^{\theta }$, so the jump product rule givesFor an explicit localization, stop at ${\tau }_n=\inf \{t:X_t\ge n\}\wedge n$ with $n>i$. Up to that time the statewise rates are bounded by the maximum of the finitely many rates below $n$, the jump count is bounded, and $M$ and its compensated-integral integrand are bounded on the finite stopped horizon. The compensated-measure theorem therefore makes each stopped process a true martingale. The times increase to the explosion lifetime (or infinity if there is no explosion), proving $M$ is a local martingale on $[0,\zeta )$ for every real $\theta$.

Now suppose $\lambda (j)\le C$ uniformly. For $C>0$, propose candidate births at the times of a rate-$C$ Poisson process, accepting each with probability $\lambda (X_{s-})/C$ using fresh independent uniform marks. While the state is fixed, Poisson thinning gives a holding time with the required exponential distribution of rate $\lambda$; accepted births have exactly the prescribed law. Thus there can be no explosion, and the true birth count $N_T$ is stochastically dominated by a Poisson random variable of mean $CT$. In particular,The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,This is one integrable dominating variable for the entire stopped family. Conditional dominated convergence removes the localization and provesso the exponential martingale of a pure birth process is a genuine martingale for uniformly bounded rates. Only uniform integrability on each finite horizon is asserted; it need not be uniformly integrable over infinite time.

For a random initial state, the same proof requires $\mathbb{E}{e}^{\theta X_0}<\infty$. The source does not explicitly specify the initial distribution, so that qualification cannot be dropped: take constant rate one and an independent initial law $\mathbb{P}(X_0=n)=6/({\pi }^2{n}^2)$. For $\theta >0$, already $\mathbb{E}{M}_0=\infty$, precluding even the integrability in the definition of a martingale. The fixed-initial-state interpretation above supplies the intended complete proof; bounded rates alone do not remedy an arbitrary heavy-tailed initial law.