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Let $(X_t)_{t\ge0}$ be a càdlàg Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ and $B\in\mathcal B([0,\infty)\times\mathbb R)$.

How can we show that $$\pi:=\sum_{\substack{s\:\ge\:0\\\Delta X_s\:\ne\:0}}1_B(s,\Delta X_s)$$ is $\mathcal A$-measurable?

$(\Delta X_s)_{s\ge0}$ is clearly $(\mathcal F_s)_{s\ge0}$-adapted and hence $$\Omega\to\{0,1\}\;,\;\;\;\omega\mapsto1_{\{\:\Delta X_s\:\ne\:0\:\}}(\omega)1_B(s,\Delta X_s(\omega))\tag1$$ is $\mathcal F_s$-measurable for all $s\ge0$. Moreover, $$\{s\ge0:\Delta X_s(\omega)\ne0\}\tag2$$ is countable for all $\omega\in\Omega$.

Now we might be tempted to argue that $\pi$ is $\mathcal A$-measurable as the countable sum of $\mathcal A$-measurable functions. However, $(2)$ is only a countable set for each fixed $\omega\in\Omega$. And in order to apply the former argument, we would need that there is a countable $D\subseteq[0,\infty)$ with $$\pi(\omega)=\sum_{s\in D}1_{\{\:\Delta X_s\:\ne\:0\:\}}(\omega)1_B(s,\Delta X_s(\omega))\;\;\;\text{for all }\omega\in\Omega.\tag3$$ Can we fix this issue?


As usual, $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for all $t\ge0$ and càdlàg $x:[0,\infty)\to\mathbb R$.

Remark: I've asked this qustion on MSE, but I didn't receive an answer (even after a bounty expired). For some reason, I cannot delete the question on MSE at the moment, but I will as soon as it is possible.

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  • $\begingroup$ Please refer to the book "J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003" Prop. II.1.16, p.69. This proposition answer your question. $\endgroup$
    – JGWang
    Jan 5, 2022 at 7:19

3 Answers 3

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This is routine (and I am quite sure covered by standard textbooks), although somewhat tedious. First, for a compactly supported, non-negative and continuous $f$, one writes $$ \tag{1} S_t[f] := \sum_{s \leqslant t} f(s, X_{s-}, X_s) = \lim_{n \to \infty} \sum_{i = 0}^{\lfloor n t\rfloor} f(\tfrac in, X_{(i-1)/n}, X_{i/n}) , $$ which shows that the left-hand side is measurable. Next, for an open and bounded $B$, one approximates the sum of $\mathbb 1_B(s, \Delta X_s)$ by $S_t[f]$ for an appropriate sequence of functions $f$. Finally, one uses the monotone class theorem, or Dynkin's lemma, to show measurability for all Borel sets $B$.


Remark. Judging by your previous questions, as well as the emphasized part of this question, you seem to be confused by the use of different $\sigma$-algebras. The above construction does not work in the framework of the product $\sigma$-algebras on the space $\mathbb R^{[0,\infty)}$ of all paths, because the set of càdlàg paths is not measurable with respect to the product $\sigma$-algebra. Instead, one works with $D([0, \infty), \mathbb R)$, the class of càdlàg paths, with the "cylindrical" $\sigma$-algebra $\mathcal A$ (which is just the previous product $\sigma$-algebra restricted to the non-measurable subset $D([0, \infty), \mathbb R)$). Finally, one augments this $\sigma$-algebra $\mathcal A$ with respect to an appropriate probability measure (or at least one considers the $\sigma$-algebra of universally measurable sets) in order to have various objects (such as hitting times) measurable.


Edit — a sketch of the proof of (1): This is a pathwise result.

We may restrict our attention to a finite time horizon: $t \in [0, T]$ for $T$ large enough, so that $f(s, x, y) = 0$ when $s \geqslant T$. Choose $\epsilon > 0$ such that $f(s, x, y) = 0$ if $|x - y| < \epsilon$, and enumerate all jumps of $X_s$ of size larger than $\tfrac\epsilon2$: these times form an increasing sequence $s_k$. Call $X_t'$ the same path as $X_t$, but with all jumps at times $s_k$ removed: $$ X_t' = X_t - \sum_{k : s_k \leqslant t} \Delta X_{s_k} . $$ Then $|\Delta X_t'| \leqslant \tfrac\epsilon2$ for all $t$. As in the proof of uniform continuity of continuous functions, one has the following property (see below for a proof): $$ \tag{2} \text{there is $\delta > 0$ such that if $|t_1 - t_2| < \delta$, then $|X_{t_1}' - X_{t_2}'| < \epsilon$.} $$ (Here, of course, $0 \leqslant t_1, t_2 \leqslant T$.)

Suppose that $\tfrac1n<\epsilon$. Then it follows that $f(\tfrac in, X_{(i-1)/n}, X_{i/n}) \ne 0$ only if the interval $[\tfrac{i-1}n, \tfrac in]$ contains some $s_k$ (for otherwise the increments of $X_t$ on this interval are equal to the increments of $X_t'$). The desired result follows now easily by continuity of $f$.


Proof of (2): Suppose, contrary to our claim, that no such $\delta$ exists, and let $p_n, q_n$ satisfy $|p_n - q_n| < \tfrac1n$ and $|X_{p_n}' - X_{q_n}'| \geqslant \epsilon$. By passing to a subsequence, we may assume that $p_n$ and $q_n$ converge to some limit $s$, and it follows that necessarily $|\Delta X_s'| \geqslant \epsilon$.

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  • $\begingroup$ Thank you for your answer. I've actually didn't found this statement in a lecture book. Do you've got a reference in mind? I've actually struggle to prove why your identity is correct and asked for it on MSE: math.stackexchange.com/q/4348849/47771. (BTW, I guess you've intended to write $\lim_{n \to \infty} \sum_{i = 0}^{\lfloor n t\rfloor} f(X_{(i-1)/n}\color{red}-X_{i/n})$ on the right-hand side.) $\endgroup$
    – 0xbadf00d
    Jan 5, 2022 at 10:57
  • $\begingroup$ Have you seen Lemma 20.9, and the proof of the Lévy–Itô decompositon that follows after it, in Sato's book? This is perhaps not the very same statement, but it contains what you are looking for (with a different argument, perhaps). Also, I corrected the typo, but it is different from what you suggested: $f$ is a function of three variables. $\endgroup$ Jan 5, 2022 at 11:54
  • $\begingroup$ Thank you for the reference. I've seen this in Sato's book before, but I would actually prefer to understand why your identity is true. Could you take a look at my MSE post in my first comment? $\endgroup$
    – 0xbadf00d
    Jan 5, 2022 at 11:59
  • $\begingroup$ Please refer to the book "J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003" Prop. II.1.16, p.69. This proposition answer your question. $\endgroup$
    – JGWang
    Jan 6, 2022 at 3:20
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    $\begingroup$ Come on: we have $\limsup_{p,q\to s} |X_p-X_q| \geqslant \epsilon$, and so necessarily $\Delta X_s| \geqslant \epsilon$. $\endgroup$ Jan 7, 2022 at 8:51
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This is just a complement of above comment.

To discuss the measurability of $\pi$, the following conclusion is helpful(cf. C. Dellacherie & P. Meyer, Probabilities and Potential, volume 29 of North-Holland Mathematics Studies. North-Holland, Amsterdam, 1978. Theorem VI.88B, p.140).

Theorem Let $X=\{X_t,t\ge0\}$ be a real valued, càdlàg adapted process, $X_{0-} =X_0$, and $D=\{(\omega,t): X_{t-}(\omega)\ne X_t(\omega)\}$. Then $D$ is a countable union of disjoint graph of stopping times, i.e., \begin{equation*} D=\bigcup_{n\ge 1}[\hspace{-1pt}[ T_n]\hspace{-1pt}]. \tag{1} \end{equation*}

One fact should be mention is that, the conclusion is no matter the completeness of $\mathscr{A},(\mathscr{F}_t)_{t\ge 0} $. Hence, $T_n$ is also $\mathscr{A}$-measurale.

Using (1), rewrite $\pi$ as following: \begin{align*} \pi(\omega,B)&= \sum_{s>0}1_{\{\Delta X_s\ne 0\}}(\omega)1_B(s,\Delta X_s(\omega))\\ &=\sum_{n\ge1} 1_B(T_n(\omega), \Delta X_{T_n(\omega)}(\omega) ). \tag{2} \end{align*}

Due to the $X$ is càdlàg adapted process, $X$ and $X_-=(X_{t-}, t\ge 0), \Delta X$ are measurable processes, i.e., $X$ and $X_-, \Delta X$ are $\mathscr{A}\otimes\mathscr{B}(\mathbb{R}_+)$-measurable. Furthermore, $\Delta X_{T_n(\omega)}(\omega), 1_B(T_n(\omega), \Delta X_{T_n(\omega)}(\omega))$ and $\pi(\omega,B)$, all are $\mathscr{A}$-measurable.

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(Assuming that the process $(X_t)$ is adapted to the filtration $(\mathcal F_t)$.)

Because $X$ is cadlag, the set $J_n(\omega):=\{t>0: |\Delta X_t(\omega)|>1/n\}$ is locally finite for each $\omega$ and each $n\ge 1$. It can be "exhausted" by a sequence of stopping times $\{T_{nk}\}_{k=1}^\infty$ defined by $T_{n0}:=0$ and $$ T_{n,k+1}:=\inf\{t>T_{nk}: \Delta |X_t|>1/n\},\qquad k=0,1,\ldots. $$ For a Borel set $B$, and each $n$ and $k$, the function $$ \pi_{nk}:\omega\mapsto 1_B(T_{nk}(\omega),\Delta X_{T_{nk}}(\omega))1_{\{T_{nk}(\omega)<\infty\}} $$ is $\mathcal A$ measurable, hence so is the countable sum $$ \pi_n:=\sum_{k\ge 1}\pi_{nk}1_{\{T_{nk}<\infty\}}. $$ Finally, $\pi_n$ increases pointwise to $\pi$ as $n\to\infty$, so $\pi$ is also $\mathcal A$ measurable.

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