= Solution
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 by
$$
Qf(x)=\int_E[f(y)-f(x)]q(x,dy).
$$
The <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 is
$$
\nu(ds,dy)=q(X_{s-},dy)ds.
$$
The 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 <martingale>
$$
(H*(\mu-\nu))_t=\int_{(0,t]\times E}H(s,y)(\mu-\nu)(ds,dy).
$$
The same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $\mathbf1_A\mathbf1_{(s,t]}H$, 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 gives
$$
\boxed{dM_t=aM_{t-}\bigl(dN_t-\lambda(X_{t-})dt\bigr),\qquad M_0=e^{\theta i}.}
$$
For 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 \b[$M$ 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,
$$
\mathbb E e^{bN_T}\le\exp\{CT(e^b-1)\}<\infty\qquad(b\ge0).
$$
The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,
$$
0\le M_{s\wedge\tau_n}
\le\exp\{\theta i+\theta_+N_T+(1-e^\theta)_+CT\}.
$$
This is one integrable dominating variable for the entire stopped family. Conditional <dominated convergence> removes the localization and proves
$$
\boxed{\mathbb E[M_t\mid\mathcal F_s]=M_s\quad(s\le t),}
$$
so 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 $\mathbb P(X_0=n)=6/(\pi^2n^2)$. For $\theta>0$, already $\mathbb EM_0=\infty$, 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.
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