= Solution
The PDF defines multiplication by an indicator: $X_t=U\mathbf1_{\{V\geq e^{-t}\}}$. Put
$$
\boxed{T=-\log V=\inf\{t\geq0:X_t>0\}.}
$$
Then $T>0$ almost surely, $\{T\leq t\}=\{X_t>0\}\in\mathcal F_t$, and $T$ is a <stopping time>. It has <exponential distribution> of rate one and is independent of $U$. The path stays at zero until $T$, jumps to $U$ at $T$, and remains there.
Let $p=\mathbb P(U\in B)$ and $N_t=\mathbf1_{\{T\leq t,U\in B\}}$. We verify the martingale property directly. On $\{T\leq s\}$ both increments $N_t-N_s$ and $(T\wedge t)-(T\wedge s)$ vanish. On $\{T>s\}$, every observation up to $s$ is zero, so the past reveals only that survival event. Conditional on it, $R=T-s$ is rate-one exponential by memorylessness and remains independent of the mark $U$. With $h=t-s$,
$$
\mathbb E[N_t-N_s\mid\mathcal F_s]=\mathbf1_{\{T>s\}}p(1-e^{-h}),
$$
while
$$
\mathbb E[(T\wedge t)-(T\wedge s)\mid\mathcal F_s]=\mathbf1_{\{T>s\}}\mathbb E(R\wedge h)=\mathbf1_{\{T>s\}}\int_0^he^{-r}dr.
$$
Subtracting $p$ times the second identity from the first gives zero. The process is adapted and integrable, since its absolute value is bounded by $1+pt$ on a finite horizon. Therefore
$$
\boxed{N_t-p(T\wedge t)\text{ is a martingale}.}
$$
On $\{T\leq t\}$ we have $X_t=U$, so this is exactly the requested process. This first-principles argument uses the natural filtration, which does not disclose the future mark or lifetime. It describes a <single marked exponential jump process>.
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