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.