Let $E$ be a normed $\mathbb R$-vector space and $x:[0,\infty)\to E$ be regular in the sense that $$x(t\pm):=\lim_{s\to t\pm}x(s)$$ exist for all $t\ge0$. Let $$\Delta x(t):=x(t+)-x(t-)\;\;\;\text{for }t\ge0$$ and $B\in\mathcal B(E)$ with $0\not\in\overline B$.
Let $b>a\ge0$, $$\mathcal D_{[a,\:b]}:=\{(t_0,\ldots,t_k):k\in\mathbb N\text{ and }a=t_0<\cdots<t_k=b\}$$ and $$|\varsigma|:=\max_{1\le i\le k}(t_i-t_{i-1})\;\;\;\text{for }k\in\mathbb N\text{ and }\varsigma=(t_0,\ldots,t_k)\in \mathcal D_{[a,\:b]}.$$
Now let $F$ be a normed $\mathbb R$-vector space, $f:(a,b]\times E\times E\to F$ with $$f(t,y_1,y_2)=0\;\;\;\text{for all }(t,y_1,y_2)\in(a,b]\times E\times E\text{ with }\|y_1-y_2\|_E<r\tag1$$ for some $r>0$ and $$S_\varsigma:=\sum_{i=1}^kf(t_i,x\left(t_{i-1}\right),x\left(t_i\right)$$ for $k\in\mathbb N$ and $\varsigma=(t_0,\ldots,t_k)\in\mathcal D_{[a,\:b]}$.
We can show that $$S_\varsigma\xrightarrow{|\varsigma|\to0+}\sum_{t\in(a,\:b]}f(t,x(t-),x(t+)).\tag2$$
Now let $(X_t)_{t\ge0}$ be an $E$-valued càdlàg process on a measurable space $(\Omega,\mathcal A)$ and $B\in\mathcal B([0,\infty)\times E)$. Using $(2)$, we can immediately conclude that $\sum_{t\in(a,\:b]}f(t,X_{t-},X_{t+})$ is $\mathcal A$-measurable.
Using $(2)$ I would like to conclude that $\sum_{\substack{s\ge0\\\Delta X_s\ne0}}1_B(s,\Delta X_s)$ is $\mathcal A$-measurable. Can we do this by approximating $1_B$ in a suitable way by suitable functions $f$?
I'm only able to prove this under the assumption that $B=(a,b]\times A$ for some $b>a\ge0$ and an open $A\subseteq E$ with $0\not\in A$ and $A\ne E$. Under this assumption, $$A_n:=\left\{x\in E:\operatorname{dist}(x,E\setminus A)\ge\frac1n\right\}$$ is closed for all $n\in\mathbb N$, $(A_n)_{n\in\mathbb N}$ is nondecreasing and $$\bigcup_{n\in\mathbb N}A_n=A.\tag3$$ We can find $(f_n)_{n\in\mathbb N}\subseteq C_b(E)$ with $0\le f_n\le 1$ and \begin{align}\left.f_n\right|_{A_n}&=1;\tag4\\\left.f_n\right|_{E\setminus A}&=0\tag5\end{align} for all $n\in\mathbb N$. It clearly holds $$f_n\xrightarrow{n\to\infty}1_A\tag6.$$ Now note that $$\sum_{\substack{s\ge0\\\Delta X_s\ne0}}1_A(s,\Delta X_s)=\sum_{s\in(a,\:b]}1_A(\Delta X_s)=\lim_{n\to\infty}\sum_{s\in(a,\:b]}f_n(\Delta X_s)\tag7$$ (since $0\not\in A$, the sum $\sum_{s\in(a,\:b]}1_A(\Delta X_s)$ is finite) and hence we should be able to conclude by $(2)$. I say should, cause I'm unsure whether $0\not\in\overline A$ really implies $0\not\in\operatorname{supp}f_n$ for all $n\in\mathbb N$.
Can we obtain the general claim from this?