Correction to: Cylindrical Martingale Problems Associated with Lévy Generators

1 Correction to: J Theor Probab (2019) 32:1306–13 https://doi.org/10.1007/s10959-018-0814-4

2 Corrections

In this note, we correct claims made in [2]:

• (i) It is claimed that the generalized martingale problem introduced in [2] allows explosion in a continuous manner. However, because the cemetery $\Delta$ is added to $\mathbb {B}$ as an isolated point, explosion can only happen by a jump and is excluded by [2, Lemma 4.3]. In Sect. 2, we explain how the setup can be adjusted to include the possibility of explosion.

• (ii) In the proof of [2, Proposition 4.8], it is needed that the operator $A$ has a non-empty resolvent set $\rho (A)$, i.e., that

  $$\begin{aligned} \rho (A) \triangleq \big \{ \lambda \in \mathbb {R}:(\lambda - A)^{-1} \text { exists in } L (\mathbb {B}, \mathbb {B})\big \} \not = \emptyset . \end{aligned}$$

  This assumption is missing in [2]. It is, e.g., satisfied in case $A$ is the generator of a $C_0$-semigroup; see [4, Remark 1.1.3, Proposition 1.2.1].

3 A Setup Including Explosion

3.1 Modified Setup

In the following, we explain how $\Omega , \tau _n$ and $\tau _\Delta$ have to be redefined such that the setting includes the possibility of explosion.

For a function $\omega :\mathbb {R}_+ \rightarrow \mathbb {B}_\Delta$, we define

$$\begin{aligned} \tau _\Delta (\omega ) \triangleq \inf (t \in \mathbb {R}_+:\omega (t) = \Delta ), \end{aligned}$$

where, as always, $\inf (\emptyset ) \triangleq \infty$. Let $\Omega$ to be the space of all right continuous functions $\omega :\mathbb {R}_+ \rightarrow \mathbb {B}_\Delta$ which are càdlàg on $[0, \tau _\Delta (\omega ))$ and satisfy $\omega (t) = \Delta$ for all $t \ge \tau _\Delta (\omega )$. The difference in comparison with the setting in [2] is that $\omega \in \{\tau _\Delta < \infty \}$ might not have a left limit at $\tau _\Delta (\omega )$.

Denote by $X$ the coordinate process, i.e., $X_t(\omega ) = \omega (t)$ for all $\omega \in \Omega$ and $t \in \mathbb {R}_+$, and denote by $\mathcal {F} \triangleq \sigma (X_t, t \in \mathbb {R}_+)$ the $\sigma$-field generated by $X$. The proof of the following is given in Sect. 2.2.

Lemma 1

There exists a metric $d_\Omega$ on $\Omega$ such that $(\Omega , d_\Omega )$ is separable and complete and $\mathcal {F}$ is the corresponding Borel $\sigma$-field.

Let $\mathbf {F}= (\mathcal {F}_t)_{t \ge 0}$ be the filtration generated by $X$, i.e. $\mathcal {F}_t \triangleq \sigma (X_s, s \in [0, t])$ for $t \in \mathbb {R}_+$. Note that $\tau _\Delta$ is an $\mathbf {F}$-stopping time, because $\{\tau _\Delta \le t\} = \{X_t = \Delta \}\in \mathcal {F}_t$. For $\Gamma \subseteq \mathbb {B}$, we define

$$\begin{aligned} \tau (\Gamma ) \triangleq \inf \big (t < \tau _\Delta :X_t \in \Gamma \ \text {or}\ X_{t-} \in \Gamma \big )\wedge \tau _\Delta . \end{aligned}$$

The proof of the following is given in Sect. 2.3.

Lemma 2

• (i) If $\Gamma \subseteq \mathbb {B}$ is closed, then $\tau (\Gamma )$ is an $\mathbf {F}$-stopping time.

• (ii) If $\Gamma _1 \subseteq \Gamma _2 \subseteq \Gamma _3 \subseteq \cdots$ is an increasing sequence of open sets in $\mathbb {B}$ such that $\bigcup _{n \in \mathbb {N}} \Gamma _n = \mathbb {B}$, then $\tau (\mathbb {B}\backslash \Gamma _n) \nearrow \tau _\Delta$ as $n \rightarrow \infty$.

We define

$$\begin{aligned} \tau _n \triangleq \inf \big (t < \tau _\Delta :\Vert X_t\Vert \ge n\ \text {or}\ \Vert X_{t-}\Vert \ge n\big ) \wedge \tau _\Delta \wedge n, \quad n \in \mathbb {N}. \end{aligned}$$

By Lemma 2, $(\tau _n)_{n \in \mathbb {N}}$ is a sequence of $\mathbf {F}$-stopping times satisfying $\tau _n \nearrow \tau _\Delta$ as $n \rightarrow \infty$. In this modified setting, the GMP can be defined as in [2] and all results from [2] hold. In Sect. 3, we comment on necessary changes in the proofs.

3.2 Proof of Lemma 1

We adapt the proof of [1, Lemma A.7]. Define

$$\begin{aligned} \Omega ^\star \triangleq \big (D (\mathbb {R}_+, \mathbb {B}) \times (0, \infty ]\big ) \cup \big (\{\omega _\Delta \} \times \{0\}\big ), \end{aligned}$$

where $\omega _\Delta (t) = \Delta$ for all $t \in \mathbb {R}_+$. For $z \in [0, \infty ]$ and $t \in \mathbb {R}_+$, we define

$$\begin{aligned} \phi _z (t)&\triangleq {\left\{ \begin{array}{ll} t,&{}z = \infty ,\\ z \big (1 - e^{- t}\big ),&{} z \in (0, \infty ),\\ 0,&{} z = 0, \end{array}\right. }\\ \phi ^{-1}_z (t)&\triangleq {\left\{ \begin{array}{ll} t,&{} z = \infty , \\ - \log \left( 1 - \frac{t}{z}\right) \mathbf {1}_{\{t < z\}},&{} z \in (0, \infty ),\\ 0,&{} z = 0. \end{array}\right. } \end{aligned}$$

Moreover, we define $\iota :\Omega \rightarrow \Omega ^\star$ by

$$\begin{aligned} \iota (\omega ) \triangleq (\omega \circ \phi _{\tau _\Delta (\omega )}, \tau _\Delta (\omega )). \end{aligned}$$

Lemma 3

$\iota$ is a bijection.

Proof

To check the injectivity, let $\omega , \alpha \in \Omega$ be such that $\iota (\omega ) = \iota (\alpha )$. In case $\tau _\Delta (\omega ) = \tau _\Delta (\alpha ) \in \{0, \infty \}$, we clearly have $\omega = \alpha$. In case $0< \tau _\Delta (\omega ) = \tau _\Delta (\alpha ) < \infty$, we can deduce from the first coordinates of $\iota (\omega )$ and $\iota (\alpha )$ that $\omega = \alpha$ on $[0, \tau _\Delta (\omega )) = [0, \tau _\Delta (\alpha ))$, which implies $\omega = \alpha$.

To check the surjectivity, note that $\iota (\omega _\Delta ) = (\omega _\Delta , 0)$ and that $\iota ( \omega \circ \phi ^{-1}_{t} \mathbf {1}_{[0, t)} + \Delta \mathbf {1}_{[t, \infty )}) = (\omega , t)$ for all $(\omega , t) \in D(\mathbb {R}_+, \mathbb {B}) \times (0, \infty ]$. $\square$

Let $d_D$ be the Skorokhod metric on $D(\mathbb {R}_+, \mathbb {B}_\Delta )$ and let $d_{[0, \infty ]}$ be the arctan metric on $[0, \infty ]$. We define

$$\begin{aligned} d_{D \times [0, \infty ]} ((\omega , t),(\alpha , s)) \triangleq d_D(\omega , \alpha ) + d_{[0, \infty ]} (t, s) \end{aligned}$$

for $(\omega , t), (\alpha , s) \in D(\mathbb {R}_+, \mathbb {B}_\Delta ) \times [0, \infty ],$ and set

$$\begin{aligned} d_{\Omega ^\star } \triangleq d_{D \times [0, \infty ]} \big |_{\Omega ^\star \times \Omega ^\star }. \end{aligned}$$

We note that $(\Omega ^\star , d_{\Omega ^\star })$ is separable and complete, because it is a $G_\delta$ subspace of $(D(\mathbb {R}_+, \mathbb {B}_\Delta ) \times [0, \infty ], d_{D \times [0, \infty ]})$. Due to Lemma 3, we can equip $\Omega$ with the metric

$$\begin{aligned} d_\Omega (\omega , \alpha )&\triangleq d_{\Omega ^\star } (\iota (\omega ), \iota (\alpha )) \\&= d_D(\omega \circ \phi _{\tau _\Delta (\omega )}, \alpha \circ \phi _{\tau _\Delta (\alpha )}) + d_{[0, \infty ]} (\tau _\Delta (\omega ), \tau _\Delta (\alpha )) \end{aligned}$$

for $\omega , \alpha \in \Omega .$ In this case, $\iota$ is an isometry and $(\Omega , d_\Omega )$ is separable and complete. In the following, we equip $\Omega$ with the topology induced by the metric $d_{\Omega }$.

We now prove that $\mathcal {F} = \mathcal {B}(\Omega )$. By the definition of the metric $d_\Omega$, the maps

$$\begin{aligned} \Omega \ni \omega \mapsto \omega \circ \phi _{\tau _\Delta (\omega )} \in D(\mathbb {R}_+, \mathbb {B}_\Delta ), \quad \Omega \ni \omega \mapsto \tau _\Delta (\omega ) \in [0, \infty ] \end{aligned}$$

are continuous. For fixed $t \in \mathbb {R}_+$, the map $[0, \infty ] \ni z \mapsto \phi ^{-1}_z(t) \in \mathbb {R}_+$ is Borel and, consequently, also

$$\begin{aligned} \Omega \ni \omega \mapsto \phi ^{-1}_{\tau _\Delta (\omega )} (t) \in \mathbb {R}_+ \end{aligned}$$

is Borel. Because right continuous adapted processes are progressively measurable, the map

$$\begin{aligned} D(\mathbb {R}_+, \mathbb {B}_\Delta ) \times \mathbb {R}_+ \ni (\omega , t) \mapsto \omega (t) \triangleq Y(\omega , t) \in \mathbb {B}_\Delta \end{aligned}$$

is Borel. We conclude that for every $t \in \mathbb {R}_+$ the map

$$\begin{aligned} \Omega \ni \omega \mapsto \omega (t) = Y(\omega \circ \phi _{\tau _\Delta (\omega )}, \phi ^{-1}_{\tau _\Delta (\omega )} (t)) \mathbf {1}_{\{t < \tau _\Delta (\omega )\}} + \Delta \mathbf {1}_{\{t \ge \tau _\Delta (\omega )\}} \in \Omega \end{aligned}$$

is Borel. This implies that $\mathcal {F} \subseteq \mathcal {B}(\Omega )$.

Note that $\iota$ is $\mathcal {F}/\mathcal {B}(\Omega ^\star )$ measurable. Let $f :\Omega \rightarrow \mathbb {R}$ be a Borel function. Because $\iota$ is an isometry, the inverse map $\iota ^{-1} :\Omega ^\star \rightarrow \Omega$ is continuous and therefore Borel. We conclude that

$$\begin{aligned} \Omega \ni \omega \mapsto f(\omega ) = ((f \circ \iota ^{-1}) \circ \iota )(\omega ) \in \mathbb {R} \end{aligned}$$

is $\mathcal {F}/\mathcal {B}(\mathbb {R})$ measurable as composition of the $\mathcal {B}(\Omega ^\star )/\mathcal {B}(\mathbb {R})$ measurable map $f \circ \iota ^{-1}$ and the $\mathcal {F}/\mathcal {B}(\Omega ^\star )$ measurable map $\iota$. This implies $\mathcal {B}(\Omega ) \subseteq \mathcal {F}$ and the proof is complete. $\square$

3.3 Proof of Lemma 2

(i). We have to show that $\{\tau (\Gamma ) \le t\} \in \mathcal {F}_t$ for all $t \in \mathbb {R}_+$. For $x \in \mathbb {B}$, we define $d (x, \Gamma ) \triangleq \inf _{y \in \Gamma } \Vert x - y\Vert$ and set

$$\begin{aligned} \Gamma _n \triangleq \big \{x \in \mathbb {B} :d(x, \Gamma ) < \tfrac{1}{n}\big \}. \end{aligned}$$

Moreover, on $\{t < \tau _\Delta \}$ we set

$$\begin{aligned} F_t \triangleq cl _{\mathbb {B}} (\{X_s :s \in [0, t]\}) = \{X_s, X_{s-} :s \in [0, t]\} \subseteq \mathbb {B}. \end{aligned}$$

Because $x \mapsto d(x, \Gamma )$ is Lipschitz continuous, the set $\Gamma _n$ is open, and because $\Gamma$ is closed, $\Gamma = \{x \in \mathbb {B} :d(x, \Gamma ) = 0\}$. Define $\tau \triangleq \sup _{n \in \mathbb {N}} \tau (\Gamma _n)$. Because $\Gamma \subseteq \Gamma _n$, it is clear that $\tau \le \tau (\Gamma )$. Next, we show that $\tau \ge \tau (\Gamma )$. We claim that this inequality follows if we show that

$$\begin{aligned} \forall t \in \mathbb {R}_+ :\bigcap _{n \in \mathbb {N}} \{F_t \cap \Gamma _n \not = \emptyset \} \subseteq \{F_t \cap \Gamma \not = \emptyset \} \text { on } \{t < \tau _\Delta \}. \end{aligned}$$

(2.1)

We explain this: In case $\tau \ge \tau _\Delta$, we have $\tau = \tau (\Gamma ) = \tau _\Delta$. Take $\omega \in \{\tau < \tau _\Delta \}$ and let $\varepsilon > 0$ be such that $\varepsilon < \tau _\Delta (\omega ) - \tau (\omega )$ in case $\tau _\Delta (\omega ) < \infty$. For each $n \in \mathbb {N}$, we find a $t_n \in [\tau (\Gamma _n) (\omega ), \tau (\Gamma _n) (\omega ) + \varepsilon )$ such that $F_{t_n} (\omega ) \cap \Gamma _n \not = \emptyset$. Note that $t \triangleq \sup _{n \in \mathbb {N}} t_n \le \tau (\omega ) + \varepsilon < \tau _\Delta (\omega )$ and that $F_t (\omega ) \cap \Gamma _n \not = \emptyset \text { for all } n \in \mathbb {N}.$ Consequently, in case (2.1) holds we have $F_t(\omega ) \cap \Gamma \not = \emptyset$, which implies $\tau (\Gamma ) (\omega ) \le t \le \tau (\omega ) + \varepsilon$. We conclude that $\tau \ge \tau (\Gamma )$ as claimed. We proceed showing (2.1). Fix $t \in \mathbb {R}_+$. Because on $\{t < \tau _\Delta \}$

$$\begin{aligned} \bigcap _{n \in \mathbb {N}} \{F_t \cap \Gamma _n \not = \emptyset \} \subseteq \big \{\inf _{x \in F_t} d(x, \Gamma ) = 0\big \}, \end{aligned}$$

it suffices to show that on $\{t < \tau _\Delta \}$

$$\begin{aligned} \big \{\inf _{x \in F_t} d(x, \Gamma ) = 0\big \} \subseteq \{F_t \cap \Gamma \not = \emptyset \}. \end{aligned}$$

Take $\omega \in \{t < \tau _\Delta \}$. Because $\{\omega (\cdot \wedge t)\}$ is compact in $D(\mathbb {R}_+, \mathbb {B})$, $F_t(\omega )$ is compact in $\mathbb {B}$ by [4, Problem 16, p. 152]. Consequently, due to its continuity, the function $x \mapsto d(x, \Gamma )$ attains its infimum on $F_t(\omega )$. Thus, because $\Gamma = \{x \in \mathbb {B} :d(x, \Gamma ) = 0\}$, if $\inf _{x \in F_t(\omega )} d(x, \Gamma ) = 0$, we have $F_t(\omega ) \cap \Gamma \not = \emptyset$. We conclude that (2.1) holds and hence that $\tau = \tau (\Gamma )$.

From the equality $\tau = \tau (\Gamma )$, we deduce that for all $t \in \mathbb {R}_+$

$$\begin{aligned} \{\tau (\Gamma ) \le t\} = \bigcap _{n \in \mathbb {N}} \{\tau (\Gamma _n) \le t\}. \end{aligned}$$

(2.2)

Fix $t \in \mathbb {R}_+$ and set $\mathbb {Q}_+^t \triangleq ([0, t) \cap \mathbb {Q}_+) \cup \{t\}$. We note that

$$\begin{aligned} \begin{aligned} \{\tau (\Gamma _{n + 1}) \le t< \tau _\Delta \}&= \bigcap _{m \in \mathbb {N}} \{\tau (\Gamma _{n + 1})< t + \tfrac{1}{m} \le \tau _\Delta \} \\&\supseteq \Big ( \bigcup _{s \in \mathbb {Q}_+^t} \{X_s \in \Gamma _{n + 1}\} \Big ) \cap \{t < \tau _\Delta \}. \end{aligned} \end{aligned}$$

(2.3)

Because $\Gamma _{n + 1}$ is open, we have

$$\begin{aligned} \tau (\Gamma _{n + 1}) = \inf \big (t < \tau _\Delta :X_t \in \Gamma _{n + 1}\big ) \wedge \tau _\Delta . \end{aligned}$$

Thus, in case $\tau (\Gamma _{n + 1}) \le t < \tau _\Delta$, the right continuity of $X$ yields that $X_{\tau (\Gamma _{n + 1})} \in cl _{\mathbb {B}}(\Gamma _{n + 1}) \subseteq \Gamma _n.$ We conclude that on $\{t < \tau _\Delta \}$

$$\begin{aligned} \{\tau (\Gamma _{n + 1}) \le t\} \subseteq \bigcup _{s \in [0, t]} \{X_s \in cl _\mathbb {B}(\Gamma _{n + 1})\} \subseteq \bigcup _{s \in \mathbb {Q}_+^t} \{X_s \in \Gamma _n\}. \end{aligned}$$

(2.4)

Now, (2.2), (2.3) and (2.4) imply that

$$\begin{aligned} \{\tau (\Gamma ) \le t < \tau _\Delta \} = \Big (\bigcap _{n \in \mathbb {N}} \bigcup _{s \in \mathbb {Q}^t_+} \{X_s \in \Gamma _{n}\} \Big ) \cap \{X_t \not = \Delta \} \in \mathcal {F}_t. \end{aligned}$$

Because

$$\begin{aligned} \{\tau (\Gamma ) \le t, \tau _\Delta \le t\} = \{\tau _\Delta \le t\} = \{X_t = \Delta \} \in \mathcal {F}_t, \end{aligned}$$

we conclude that $\tau (\Gamma )$ is a stopping time. The proof of (i) is complete.

(ii). Because $n \mapsto \tau (\mathbb {B} \backslash \Gamma _n)$ is increasing, $\tau (\mathbb {B} \backslash \Gamma _n) \nearrow \tau \triangleq \sup _{n \in \mathbb {N}} \tau (\mathbb {B}\backslash \Gamma _n)$ as $n \rightarrow \infty$. Because $\tau \le \tau _\Delta$, it suffices to show that $\tau \ge \tau _\Delta$. For contradiction, suppose that there exists an $\omega \in \{\tau < \tau _\Delta \}$ and set $\omega ' \triangleq \omega (\cdot \wedge \tau (\omega )) \in D(\mathbb {R}_+, \mathbb {B})$. Then,

$$\begin{aligned} \tau (\mathbb {B} \backslash \Gamma _n) (\omega ') = \inf \big (t \in \mathbb {R}_+ :\omega '(t) \not \in \Gamma _n \text { or } \omega '(t-) \not \in \Gamma _n \big ) \nearrow \infty \text { as } n \rightarrow \infty . \end{aligned}$$

Because $\tau (\mathbb {B}\backslash \Gamma _n)$ is an $\mathbf {F}$-stopping time by (i), so is $\tau$ and Galmarino’s test (see [6, Lemma III.2.43]) implies that $\tau (\omega ) = \tau (\omega ') = \infty$. This is a contradiction and $\tau = \tau _\Delta$ follows. The proof of (ii) is complete. $\square$

4 Modifications, Corrections and Comments on Proofs

4.1 [2, Lemma 4.3]

The last conclusion in [2, Lemma 4.3] is empty: In the setting of [2], it cannot happen that $X_{\tau _\Delta -} = \Delta$.

4.2 [2, Lemmata 4.3, 4.5]

Due to the initial value and the possibility that $X$ has no left limit at $\tau _n$, some bounds in the proofs of [2, Lemmata 4.3, 4.5] are only valid on the open stochastic interval $]]0, \tau _n[[$. Because singletons have Lebesgue measure zero, the arguments require no further changes.

The last conclusion in the proof of [2, Lemma 4.5] follows from the dominated convergence theorem.

4.3 [2, Proposition 4.8]

In the proof, it has been used that $\rho (A^*) \not = \emptyset$, see [8, Lemma 4.1]. Because $\mathbb {B}$ is separable and reflexive, its dual $\mathbb {B}^*$ is separable and $D$ in the proof of [2, Proposition 4.8] can be constructed more directly: The assumption $\rho (A) \not = \emptyset$ implies that $\rho (A^*) \not = \emptyset$, see [7, Theorem 5.30, p. 169]. Let $D' \subset \mathbb {B}^*$ be a countable dense subset of $\mathbb {B}^*$ and take $\lambda \in \rho (A^*)$. Now, set $R(\lambda , A^*) \triangleq (\lambda - A^*)^{-1}$ and define $D \triangleq \{R(\lambda , A^*) x :x \in D' \} \subseteq D(A^*)$. We claim that for each $x \in D(A^*)$ there exists a sequence $(x_n)_{n \in \mathbb {N}} \subset D$ such that $x_n \rightarrow x$ and $A^* x_n \rightarrow A^* x$ in the operator norm as $n \rightarrow \infty$. To see this, take $x \in D(A^*)$ and set $y \triangleq \lambda x + A^* x$. There exists a sequence $(y_n)_{n \in \mathbb {N}}\subset D'$ such that $y_n \rightarrow y$ as $n \rightarrow \infty$. Finally, set $x_n \triangleq R(\lambda , A^*) y_n \in D$. Because $R(\lambda , A^*)\in L(\mathbb {B}^*, \mathbb {B}^*)$, we have $x_n \rightarrow R(\lambda , A^*) y = x$ as $n \rightarrow \infty$. Moreover, the triangle inequality yields that

$$\begin{aligned} \Vert A^* x_n - A^* x\Vert \le \Vert y_n - y\Vert + |\lambda | \Vert x_n - x\Vert \rightarrow 0 \text { as } n \rightarrow \infty . \end{aligned}$$

The claim is shown.

4.4 [2, Lemma 4.10]

Due to Lemma 1, it is not necessary to pass to $D(\mathbb {R}_+, \mathbb {B}_\Delta )$. Moreover, it can be seen more easily that $\Phi$ is Borel. Indeed, $\Phi$ is continuous.

4.5 [2, Lemma 4.11]

In the proof of $P$-a.s.

$$\begin{aligned} E^P \big [ \big (M^f_{t \wedge \tau _n} - M^f_{s \wedge \tau _n} \big ) \circ \theta _\xi \mathbf {1}_{\{\xi < \tau _\Delta \}} \big | \mathcal {F}_{s + \xi } \big ] = 0, \end{aligned}$$

the variable $n$ is used twice, which results in a conflict of notation. We correct the argument: Note that $\tau _{n + k} \circ \theta _\xi + \xi \le \tau _{2 (n + k)}$ on $\{\xi < \tau _{n + k}\}$ for all $k \in \mathbb {N}$. Set $\sigma _r \triangleq r \wedge \tau _n \circ \theta _\xi + \xi$. We obtain that $P$-a.s.

$$\begin{aligned}&E^P \big [ \big (M^f_{t \wedge \tau _n} - M^f_{s \wedge \tau _n} \big ) \circ \theta _\xi \mathbf {1}_{\{\xi< \tau _\Delta \}} \big | \mathcal {F}_{s + \xi } \big ]\\&\quad = \lim _{k \rightarrow \infty } E^P \big [ \big (M^f_{\sigma _t} - M^f_{\sigma _s} \big ) \mathbf {1}_{\{\xi< \tau _{n + k}\}} \big | \mathcal {F}_{s + \xi } \big ]\\&\quad = \lim _{k \rightarrow \infty } E^P \big [ \big (M^f_{\sigma _t \wedge \tau _{2(n + k)}} - M^f_{\sigma _s \wedge \tau _{2(n + k)}} \big ) \mathbf {1}_{\{\xi< \tau _{n + k}\}} \big | \mathcal {F}_{s + \xi } \big ]\\&\quad = \lim _{k \rightarrow \infty } \big (M^f_{\sigma _t \wedge \tau _{2(n + k)} \wedge (s + \xi )} - M^f_{\sigma _s \wedge \tau _{2(n + k)} \wedge (s + \xi )} \big ) \mathbf {1}_{\{\xi < \tau _{n + k}\}} = 0, \end{aligned}$$

by the optional stopping theorem.

4.6 [2, Section 4.3.2]

Because $X$ has no left limit at $\tau _\Delta$, the random measure $\mu ^X$ cannot be defined as in [2, Eq. 4.20]. We pass to a stopped version: Let $\widehat{X}$ be defined as in Eq. 4.11 in [2] and set $X^n \triangleq \widehat{X}_{\cdot \wedge \tau _n}$ and

$$\begin{aligned} \mu ^n(\omega ; dt, dx)&\triangleq \sum _{s > 0} \mathbf {1}_{ \{\Delta X^n_s (\omega ) \not = 0\}} \varepsilon _{(s, \Delta X^n_s (\omega ))} (dt, dx),\\ \nu ^n (\omega ; dt, dx)&\triangleq \mathbf {1}_{\{t \le \tau _n(\omega )\}} K(X^n_{t} (\omega ), dx) dt. \end{aligned}$$

We have the following version of [2, Lemmata 4.17, 4.18, 4.19]:

Lemma 4

For all $n \in \mathbb {N}$ the random measure $\mu ^n$ is $(\mathbf {F}^P, P)$-optional with $\mathscr {P}^P$-$\sigma$-finite Doléans measure and $(\mathbf {F}^P, P)$-predictable compensator $\nu ^n$.

Because the proofs of [2, Lemmata 4.17, 4.18] contain typos and the proof of [2, Lemma 4.19] requires some minor modification, as the set $\mathcal {Z}_1 \times \mathcal {Z}_2$ has not all claimed properties, we give a proof:

Proof

Due to [3, Theorem IV.88B, Remark below], the set $\{\Delta X^n \not = 0\}$ is $\mathbf {F}^P$-thin. Hence, [6, II.1.15] yields that $\mu ^n$ is $\mathbf {F}^P$-optional. It follows as in [9, Example 2, pp. 160] that $M^P_{\mu ^n}$ is $\mathscr {P}^P$-$\sigma$-finite. Next, we show that $\nu ^n$ is $\mathbf {F}^P$-predictable with $\mathscr {P}^P$-$\sigma$-finite Doléons measure $M^P_{\nu ^n}$. For $m \in \mathbb {N}$ we set $G_m \triangleq \{x \in \mathbb {B} :\Vert x\Vert \ge \tfrac{1}{m}\} \cup \{0\}.$ Let $W$ be a nonnegative $\mathscr {P}^P \otimes \mathscr {B}(\mathbb {B})$-measurable function which is bounded by a constant $c > 0$. Because $P$-a.s.

$$\begin{aligned} W \mathbf {1}_{[[0, \tau _m]]} \mathbf {1}_{G_m} \star \nu ^n_\infty \le c m \sup _{\Vert x\Vert \le m} K(x, \{z \in \mathbb {B} :\Vert z\Vert \ge \tfrac{1}{m}\}) < \infty , \end{aligned}$$

we conclude that $M^P_{\nu ^n}$ is $\mathscr {P}^P$-$\sigma$-finite. Furthermore, the process

$$\begin{aligned} W \star \nu ^n = \lim _{m \rightarrow \infty } W \mathbf {1}_{[[0, \tau _m]]} \mathbf {1}_{G_m} \star \nu ^n \end{aligned}$$

is $\mathbf {F}^P$-predictable as the pointwise limit of an $\mathbf {F}^P$-predictable process. We conclude that $\nu ^n$ is an $\mathbf {F}^P$-predictable random measure.

It remains to show that $\nu ^n$ is the $(\mathbf {F}^P, P)$-predictable compensator of $\mu ^n$. Let $\mathcal {Z}_1$ be the collection of sets $A \times \{0\}$ for $A \in \mathcal {F}^P_0$ and $[[0, \xi ]]$ for all $\mathbf {F}^P$-stopping times $\xi$, and let $\mathcal {Z}_2$ be the collection of all sets

$$\begin{aligned} G \triangleq \{x \in \mathbb {B} :(\langle x, y^*_1\rangle , \dots , \langle x, y^*_d \rangle ) \in A\} \in \mathcal {B}(\mathbb {B}), \end{aligned}$$

(3.1)

for $A \in \mathcal {B}(\mathbb {R}^d), y^*_1, \dots , y^*_d \in D(A^*)$ and $d \in \mathbb {N}$. Note that $M^P_{\mu ^n} (A \times \{0\} \times G) = M^P_{\nu ^n} (A \times \{0\} \times G) = 0$ for all $A \in \mathcal {F}^P_0$ and $G \in \mathcal {B}(\mathbb {B})$. Fix an $\mathbf {F}^P$-stopping time $\xi$ and the cylindrical set $G$ given by (3.1). Denote $Y^n \triangleq (\langle X^n, y^*_1\rangle , \dots , \langle X^n, y^*_d\rangle ).$ By [2, Lemma 4.7], we obtain

$$\begin{aligned} E^P\Big [ \mathbf {1}_{[[0, \xi ]]\times G} \star \mu ^n_\infty \Big ]&= E^P\Big [ \mathbf {1}_{[[0, \xi ]]\times A} \star \mu ^{Y^n}_\infty \Big ] = E^P \Big [\mathbf {1}_{[[0, \xi ]]\times G} \star \nu ^n_\infty \Big ], \end{aligned}$$

which implies $M^P_{\mu ^n} = M^P_{\nu ^n}$ on $\mathcal {Z}_1 \times \mathcal {Z}_2$. Take a norming sequence $(x^*_m)_{m \in \mathbb {N}}\subset \mathbb {B}^*$ of unit vectors, see p. 522 in [5] for a definition, and note that

$$\begin{aligned} B_m \triangleq \big \{ x \in \mathbb {B} :\Vert x\Vert> \tfrac{1}{m}\big \} = \bigcup _{k \in \mathbb {N}} \big \{ x \in \mathbb {B} :|\langle x, x^*_k\rangle | > \tfrac{1}{m}\big \}. \end{aligned}$$

For $m, k \in \mathbb {N}$ set

$$\begin{aligned} \gamma (m, k) \triangleq \inf (t \in \mathbb {R}_+ :\mu ^n ([0, t] \times B_m) > k) \wedge m. \end{aligned}$$

The dominated convergence theorem yields that

$$\begin{aligned} M^P_{\mu ^n} ((A \times B) \cap ([[0, \gamma (m, k)]]\times B_m)) = M^P_{\nu ^n}((A \times B) \cap ([[0, \gamma (m, k)]]\times B_m)) \end{aligned}$$

for all $A \times B \in \mathcal {Z}_1 \times \mathcal {Z}_2$. Now, we conclude from the uniqueness theorem for measures that $M^P_{\mu ^n} = M^P_{\nu ^n}$ on the trace $\sigma$-field $(\mathscr {P}^P \otimes \mathcal {B}(\mathbb {B}))\cap ([[0, \gamma (m, k)]]\times (B_m \cup \{0\}))$. Finally, taking $k, m \rightarrow \infty$ and using the monotone convergence theorem show that $M^P_{\mu ^n} = M^P_{\nu ^n}$ on $\mathscr {P}^P \otimes \mathcal {B}(\mathbb {B})$. The proof is complete. $\square$

The candidate density process $Z$ can be defined as in [2, Lemma 4.21] with $\mu ^X$ and $\nu ^X$ replaced by $\mu ^n$ and $\nu ^n$.

4.7 [2, Lemmata 4.21, 4.22]

In the proofs, the process $X$ should be replaced by $\widehat{X}$.

4.8 [2, Proposition 3.7]

The representation of the CMG densities and the function $V^k$ in [2, Lemma 4.23] should be multiplied by $\mathbf {1}_{\{\tau _n < \tau _\Delta \}}$. Moreover, in all Lebesgue integrals $X_-$ should be replaced by $X$.

4.9 [2, Lemma 3.16]

Instead of the Yamada–Watanabe argument, the uniqueness also follows from the observation that for a pseudo-contraction semigroup $(S_t)_{t \ge 0}$ and a square integrable Lévy process $L$ the law of $\int _0^\cdot S_{\cdot - s} dL_s$ is completely determined by $L$. This can be seen with the approximation argument used in the proof of [11, Theorem 9.20].

5 Final Comment

Above [2, Proposition 3.9] it is noted that “in a non-conservative setting, one can try to conclude existence from an extension argument in a larger path space, [but] in this case one has to prove that the extension is supported on $(\Omega , \mathcal {F})$” as defined in [2]. The larger path space, to which this comment refers, is the path space defined in this correction note. In our modified setting, it follows from Parthasarathy’s extension theorem (see [10]) that under the assumptions imposed in [2] the GMP $(A, b', a, K', \eta , \tau _\Delta -)$ has a solution whenever the GMP $(A, b, a, K, \eta , \tau _\Delta -)$ has a solution. This observation extends [2, Theorem 3.6].