Is the function $x\mapsto(\Delta x(t))_{t\ge0}$ measurable with respect to the product $\sigma$-algebra?

Question

Let $E$ be a normed $\mathbb R$-vector space. If $x:[0,\infty)\to E$ is càdlàg, let $$x(t-):=\lim_{s\to t-}x(s)\;\;\;\text{for }t\ge0$$ ($x(0-):=0$) and $$\Delta x(t):=x(t)-x(t-)\;\;\;\text{for }t\ge0.$$ Let $D([0,\infty),E)$ denote the set of all càdlàg $x:[0,\infty)\to E$.

Can we show that the function $$f:E^{[0,\:\infty)}\to E^{[0,\:\infty)}\;,\;\;\;x\mapsto\begin{cases}\left(\Delta x(t)\right)_{t\ge0}&\text{, if }x\text{ is càdlàg}\\0&\text{, otherwise}\end{cases}$$ is measurable with respect to the product $\sigma$-algebra $\mathcal B(E)^{\otimes[0,\:\infty)}$ on $E^{[0,\:\infty)}$?

The claim is not immediate to me, since $D([0,\infty),E)$ is not in $\mathcal B(E)^{\otimes[0,\:\infty)}$, even when $E$ is complete and/or separable (which you may feel free to assume, if this is necessary to obtain a positive answer).

Answer 1

No, $f$ is not measurable.

Let $$A = \{x \in E^{[0,\infty)} : x(1) \ne 0\},$$ and consider $$B = f^{-1}(A) = \{x \in D([0,\infty), E) : \Delta x(1) \ne 0\}.$$ It remains to repeat the proof that $D([0,\infty), E)$ is not measurable with respect to the product $\sigma$-algebra $\mathcal B(E)^{\otimes [0,\infty)}$, replacing $D([0,\infty), E)$ by $B$.

Suppose, contrary to our claim, that $B$ is measurable. Then there exists a countable set of indices $I$ such that $$ \text{if $x = y$ on $I$, then either both or none of $x$, $y$ are in $B$.} $$ With no loss of generality we may assume that $I$ is dense in $[0,\infty)$. Now just take $$x = \mathbb 1_{[1,\infty)} \quad \text{ and } \quad y = \mathbb 1_{[1,\infty) \cap I} .$$ Then $x = y$ on $I$ and $x \in B$, but $y \notin B$ (because $y$ is not càdlàg), a contradiction.