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 $B\in\mathcal B([0,\infty)\times\mathbb R)$.



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> 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?
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$(\Delta X\_s)\_{s\ge0}$ 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 $\omega\in\Omega$.



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> 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](https://math.stackexchange.com/q/4341779/47771), 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.



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


