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