= Solution
Take $R_t=0$ when no record has occurred below $t$, the needed convention for the maximum of an empty set. Assume $F(t)<1$ and write $S(t)=1-F(t)$. Given the latest record $v_k<t$, the chance that the next one exceeds $t$ is $S(t)/S(v_k)$. Multiplying this by the joint density from part (a) leaves
$$
S(t)\prod_{j=1}^k\frac{f(v_j)}{S(v_j)}
$$
on the ordered simplex $0<v_1<\cdots<v_k<t$. Its <integral> is $S(t)\Lambda(t)^k/k!$, where $\Lambda(t)=\int_0^t f(s)/S(s)\,ds=-\log S(t)$. The $k=0$ probability is directly $S(t)=e^{-\Lambda(t)}$. Hence
$$
\boxed{R_t\sim\operatorname{Poisson}(\Lambda(t)),\qquad\lambda(t)=\frac{f(t)}{1-F(t)}.}
$$
This is <Poisson record counts in cumulative hazard coordinates>; the instantaneous rate is the original distribution's <hazard function>. If the distribution has a finite upper endpoint and $t$ reaches it, $F(t)=1$ and infinitely many record values lie below $t$ almost surely; there is then no finite-parameter Poisson law. The finite-hazard assumption is thus necessary for the literal assertion.
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