Is $\sum_{\substack{s\:\ge\:0\\\Delta X_s\:\ne\:0}}1_B(s,\Delta X_s)$ measurable for fixed $B\in\mathcal B([0,\infty)\times\mathbb R)$?

Question

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?

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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.

Answer 1

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 $B$.

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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.

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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 $f$.

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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 $|\Delta X_s'| \geqslant \epsilon$.

Answer 2

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.

Answer 3

(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.