Marked waiting law from finite no-jump evolution

Each $f_{\sigma ,a}(u)$ is jointly continuous in $(\sigma ,u)$ and nonnegative. For fixed $\sigma$ the density measures of [F6] on the marked components, together with the nonnegative atom at $+\mathrm{\infty }$, define a measure on $(E,\mathcal{E})$. Indeed, ${\sum }_a{f}_{\sigma ,a}(u)=\mathrm{Tr}(Q{S}_u(\sigma ))=-s_{\sigma }^{\mathrm{\prime }}(u)$, so ${\sum }_a{\int }_0^t{f}_{\sigma ,a}(u)du=1-s_{\sigma }(t)$. The decreasing nonnegative survivor has a limit at infinity; [F7] applied to the densities restricted to $(0,n]$ gives total finite-event mass $1-s_{\sigma }(\mathrm{\infty })$. Adding the atom normalizes the measure to one. If $M$ is empty, the derivative identity gives $s_{\sigma }(t)=1$ and all mass is at $+\mathrm{\infty }$.

Subtracting the finite-event mass up to $u$ from one gives survivor $s_{\sigma }(u)$, since individual finite times have zero density-measure mass. Division by this strictly positive survivor gives the displayed marked hazard. The positive no-event matrix normalized by its trace is a density matrix, as is the positive post-event matrix whenever its trace $f_{\sigma ,a}(u)$ is positive. Thus the stated conditional-state and hazard formulas hold with no exceptional finite survival times.

Fix $A\in \mathcal{E}$ and put $g_A(\sigma ,u)={\sum }_{a\in M}{\mathbf{1}}_{A_a}(u)f_{\sigma ,a}(u)$. The spaces $D_d$ and $(0,\mathrm{\infty })$ have countable bases inherited from Euclidean spaces, so each open subset of their product is a countable union of basic open rectangles. Consequently the jointly continuous densities are measurable for the product sigma-algebra, and so is the nonnegative $g_A$. For every integer $n\ge 1$, $C_n(\sigma ,C)=\lambda (C\cap (0,n])$ is a finite kernel from $D_d$ to $(0,\mathrm{\infty })$: its section is a measure of total mass $n$ and each evaluation is constant in $\sigma$. By [F8], $G_n(\sigma )=\int g_A(\sigma ,u)C_n(\sigma ,du)$ is Borel measurable. By [F7], $G_n\uparrow G$, where $G(\sigma )={\sum }_a{\int }_{A_a}{f}_{\sigma ,a}d\lambda$. For every real $c$, $\{G>c\}={\bigcup }_n\{G_n>c\}$, so $G$ is measurable. Also $s_{\sigma }(\mathrm{\infty })={\inf }_{n\ge 1}{s}_{\sigma }(n)$ and $\{\sigma :s_{\sigma }(\mathrm{\infty })<c\}={\bigcup }_n\{\sigma :s_{\sigma }(n)<c\}$, a Borel set since every $s_{\sigma }(n)$ is continuous in $\sigma$. Hence the infinity mass is measurable. Their sum $K(\sigma ,A)$ is measurable for every $A\in \mathcal{E}$; together with step 1.1 this is exactly the probability-kernel property.

Matrix multiplication, exponentiation and trace are continuous, and $s_{\sigma }(u)>0$ everywhere on $D_d\times [0,\mathrm{\infty })$. Thus the no-event state and marked hazards are continuous. For each mark the set $\{(\sigma ,u):f_{\sigma ,a}(u)>0\}$ is open, and $J$ is continuous there and constant on its Borel complement. Hence $J$ is Borel measurable into $D_d$ with its subspace Borel sigma-algebra. For any one of these state maps $F$ and any Borel $B\subseteq D_d$, the Dirac-kernel evaluation is ${\delta }_{F(x)}(B)={\mathbf{1}}_B(F(x))$, which is measurable, and each Dirac measure has mass one. Finally the zero-density event set has probability ${\sum }_a{\int }_{\{u:f_{\sigma ,a}(u)=0\}}{f}_{\sigma ,a}(u)d\lambda (u)=0$ for every $\sigma$; assigning $I\mathrm{/}d$ there leaves the law unchanged. All formulas specify the maps simultaneously, without selecting separate conditional versions.