Step 1 We show that
\(j \mapsto \mu _j (G)\) is Borel for all
\(G \in \mathcal {D}\), following the strategy outlined in [
54, Exercise 6.7.4]. Recall that we assume that
\(A^*\) contains a countable determining set
\(\widetilde{A}\). Let
\(\mathcal {P}\) be the space of Borel probability measures on
\(D(\mathbb {R}_+, J)\) equipped with the topology of convergence in distribution. Note that a
\({\mathbf {D}}^o\)-adapted process is a
\({\mathbf {D}}\)-martingale if and only if it is a
\({\mathbf {D}}^o\)-martingale. The implication
\(\Rightarrow \) follows from the downward theorem (see [
48, Theorem II.51.1]) and the implication
\(\Leftarrow \) follows from the tower rule. For
\(g \in \widetilde{A}\) set
$$\begin{aligned} K^g_{t} \triangleq g(X_{t}) - g(X_0) - \int _0^{t} K g(X_{s}) ds, \quad t \in \mathbb {R}_+. \end{aligned}$$
Define
\(\mathcal {I}\) to be the set of all
\(\nu \in \mathcal {P}\) such that
\(\nu \circ X_0^{-1} = \delta _j\) for some
\(j \in J\). Moreover, let
\(\mathcal {M}\) be the set of all
\(\nu \in \mathcal {P}\) such that
$$\begin{aligned} {\mathbb {E}}^\nu \big [ (K^g_{t \wedge \rho _m} - K^g_{s \wedge \rho _m})\mathbb {1}_G\big ] = 0, \end{aligned}$$
for all
\(g \in \widetilde{A}\), all rational
\(s < t, m \in \mathbb {N}\) and
\(G\) in a countable determining class of
\(\mathcal {D}^o_s\). By the uniqueness assumption,
\(\{\mu _j, j \in J\} = \mathcal {I} \cap \mathcal {M}\). Because the set
\(\{\delta _j, j \in J\}\) is Borel due to [
8, Theorem 8.3.7] and
\(\nu \mapsto \nu \circ X_0^{-1}\) is continuous,
\(\mathcal {I}\) is Borel. The set
\(\mathcal {M}\) is Borel due to [
1, Theorem 15.13]. We conclude that
\(\{\mu _j, j \in J\}\) is Borel. Let
\(\Phi :\{\mu _j, j \in J\} \rightarrow J\) be defined by
\(\Phi (\mu _j) = j\) for all
\(j \in J\). We note that
\(\Phi \) is a continuous injection. Thus, the inverse map
\(\Phi ^{-1}\) is Borel due to Kuratovski’s theorem ([
8, Proposition 8.3.5]). This means that
\(j \mapsto \mu _j(G)\) is Borel for all
\(G \in \mathcal {D}\).
Step 3 For every
\(t \in \mathbb {R}_+\) we denote by
\(\theta _t :D(\mathbb {R}_+, J) \rightarrow D(\mathbb {R}_+, J)\) the shift operator given by
\(\theta _t \omega (s) = \omega (t + s)\). Recalling that
\(\rho \) is bounded, we deduce from [
30, Lemma III.2.44] that
$$\begin{aligned}\mathcal {D}^o_\rho \vee \theta ^{-1}_\rho (\mathcal {D}) = \mathcal {D}.\end{aligned}$$
Hence, we can associate to each
\(G \in \mathcal {D}\) a (not necessarily unique)
\(G' \in \mathcal {D}^o_{\rho } \otimes \mathcal {D}\) such that
$$\begin{aligned} G = \big \{\omega \in D :(\omega , \theta _{\rho (\omega )}\omega ) \in G'\big \}. \end{aligned}$$
We define
$$\begin{aligned} \nu (G) \triangleq \int \mu _n(d\omega ) \mu _{\omega (\rho (\omega ))} (d\omega ^*) \mathbb {1}_{G'} (\omega , \omega ^*). \end{aligned}$$
It follows from [
30, Lemma III.2.47] that
\(\nu \) is a probability measure, i.e. that
\(\nu \) is defined unambiguously. Our goal is to show that
\(\nu \) solves the martingale problem
\((A^*, L^*, j_0, \infty )\). Providing an intuition,
\(\nu \) is the law of
$$\begin{aligned} {\left\{ \begin{array}{ll} X^1_t,&{} t < \rho (X^1),\\ X^2_{t - \rho (X^1)}, &{}t \ge \rho (X^1), \end{array}\right. } \end{aligned}$$
in case
\(X^1\) is sampled according to
\(\mu _n\) and
\(X^2\) is sampled according to
\(\mu _{j}\) with
\(j = X^1_{\rho (X^1)}\). In other words, we extend
\(\mu _n\) to a solution of the global martingale problem. For
\(G \in \mathcal {D}^o_0\) we can choose
\(G' = G \times D (\mathbb {R}_+, J)\). Consequently,
$$\begin{aligned} \nu (X_0 = j_0) = \mu _n(X_0 = j_0) = 1. \end{aligned}$$
Let
\(\psi \) be a bounded
\({\mathbf {D}}^o\)-stopping time and fix
\(m \in \mathbb {N}\). For
\(\omega , \alpha \in D(\mathbb {R}_+, J)\) and
\(t \in \mathbb {R}_+\) we set
$$\begin{aligned} z(\omega , \alpha ) (t) \triangleq {\left\{ \begin{array}{ll} \omega (t),&{}t < \rho (\omega ),\\ \alpha (t - \rho (\omega )),&{}t \ge \rho (\omega ),\end{array}\right. } \end{aligned}$$
and
$$\begin{aligned} V(\omega , \alpha ) \triangleq {\left\{ \begin{array}{ll} \big ((\psi \wedge \rho _m) \vee \rho - \rho \big ) (z(\omega , \alpha )), &{}\alpha (0) = \omega (\rho (\omega )),\\ 0,&{}\text {otherwise}.\end{array}\right. } \end{aligned}$$
Due to [
18, Theorem IV.103] the map
\(V\) is
\(\mathcal {D}^o_\rho \otimes \mathcal {D}\)-measurable such that
\(V(\omega , \cdot )\) is a
\({\mathbf {D}}^o\)-stopping time for all
\(\omega \in D(\mathbb {R}_+, J)\). Furthermore, it is evident from the definition that
$$\begin{aligned} (\psi \wedge \rho _m) (\omega ) \vee \rho (\omega ) = \rho (\omega ) + V(\omega , \theta _{\rho (\omega )} \omega ) \end{aligned}$$
for
\(\omega \in D (\mathbb {R}_+, J)\). For all
\(\omega \in \{\rho < \psi \wedge \rho _m\} \in \mathcal {D}_\rho ^o\) and
\(\alpha \in D(\mathbb {R}_+, J)\) with
\(\alpha (0) = \omega (\rho (\omega ))\) we have
\( V(\omega , \alpha ) \le \rho _m(\alpha ). \) Note further that for
\(\omega \in \{\rho < \psi \wedge \rho _m\}\)$$\begin{aligned} K^g_{V(\omega , \theta _{\rho (\omega )} \omega )} (\theta _{\rho (\omega )}\omega )&= K^g_{(\psi \wedge \rho _m)(\omega ) - \rho (\omega )} (\theta _{\rho (\omega )} \omega ) \\&= g(\omega ((\psi \wedge \rho _m)(\omega ))) - g(\omega (\rho (\omega ))) - \int _{\rho (\omega )}^{(\psi \wedge \rho _m)(\omega )} Kg (\omega (s)) ds\\&= K^g_{(\psi \wedge \rho _m)(\omega )}(\omega ) - K^g_{\rho (\omega )}(\omega ). \end{aligned}$$
Because
\(K^g_{\cdot \wedge \rho }\) is a
\(\mu _n\)-martingale, we have
$$\begin{aligned} {\mathbb {E}}^\nu \big [ K^g_{\rho \wedge \psi \wedge \rho _m}\big ] = {\mathbb {E}}^{\mu _n} \big [K^g_{\rho \wedge \psi \wedge \rho _m}\big ] = 0, \end{aligned}$$
due to the optional stopping theorem. Therefore, we obtain
$$\begin{aligned} {\mathbb {E}}^\nu \big [ K^g_{\psi \wedge \rho _m} \big ]&= {\mathbb {E}}^\nu \big [ K^g_{\psi \wedge \rho _m} - K^g_{\rho \wedge \psi \wedge \rho _m}\big ]\\&= {\mathbb {E}}^\nu \big [ \big (K^g_{\psi \wedge \rho _m} - K^g_{\rho }\big ) \mathbb {1}_{\{\rho< \psi \wedge \rho _m\}}\big ]\\&= {\mathbb {E}}^\nu \big [K^g_{V (\cdot , \theta _\rho )} ( \theta _\rho ) \mathbb {1}_{\{\rho< \psi \wedge \rho _m\}}\big ]\\&=\int \mu _n(d\omega ) {\mathbb {E}}^{\mu _{\omega (\rho (\omega ))}} \big [ K^g_{V (\omega , \cdot ) \wedge \rho _m}\big ] \mathbb {1}_{\{\rho (\omega ) < (\psi \wedge \rho _m)(\omega )\}} = 0, \end{aligned}$$
again due to the optional stopping theorem (recall that
\(V(\omega , \cdot )\) is bounded and that
\(K^g_{\cdot \wedge \rho _m}\) is a
\(\mu _j\)-martingale for all
\(j \in J\)). We conclude from [
47, Proposition II.1.4] that
\(K^g_{\cdot \wedge \rho _m}\) is a
\(\nu \)-martingale and hence that under
\(\nu \) the coordinate process
\((X_t)_{t \ge 0}\) solves the martingale problem
\((A^*, L^*, j_0, \infty )\). The uniqueness assumption implies that
\(\nu = \mu ^*\). Because also for
\(G \in \mathcal {D}^o_\rho \) we can choose
\(G' = G \times D(\mathbb {R}_+, J)\), we obtain that
\( \mu ^* (G) = \nu (G) = \mu _n(G). \) The proof is complete.
\(\square \)