Let $D([0,T],\mathbb R)$ be the space of real-valued functions on $[0,T]$ that are right continuous and have left limits (càdlàg). Equip it with the (Kolmogorov or equivalently $J_1$) $\sigma$-algebra $\mathcal D$, which is the $\sigma$-algebra generated by the cylinder sets $$ \{f\in D([0,T],\mathbb R): f(t)\in B \},\quad t\in[0,T], \, B\in \mathcal B(\mathbb R), $$ for $\mathcal B(\mathbb R)$ the Borel $\sigma$-algebra on $\mathbb R$ (see Theorem 12.5).
Question: Is the set $$A=\{f: f(t)>0\,\,\forall \,t\in [0,T]\}\in\mathcal D\,\, ?$$
Comments: The set $A_1=\{f: f(t)\in C\,\,\forall \,t\in [0,T]\} \in\mathcal D$ for any closed $C$ because it equals $\cap_{q\in\mathbb Q\cap[0,T]}\{f: f(q)\in C\}$ using right continuity of the $f$'s. For the question above I tried to show that $A^c=\{f:\exists \,t\in[0,T] :\, f(t)\le 0\}\in\mathcal D$ because $A^c=\{f:\exists \,t\in[0,T] :\, f(t)< 0\}\cup \{f:\exists \,t\in[0,T] :\, f(t)= 0\}$ and the first set is measurable (by the argument for $A_1$). I then tried to show that the second set is measurable, but the closest I got is $$ \bigcap_{n\in\mathbb N}\bigcup_{q\in \mathbb Q\cap [0,T]}\{f:f(q)\in B_{1/n}(0) \}\\ = \{f:\exists \,t\in[0,T] :\, f(t)= 0\,\text{or}\,f(t-)=0\}. $$ Alternatively, can one use Billingsley 1968, page 232 that shows that the map $(f,t)\mapsto f(t)$ is $\mathcal D\times \mathcal B([0,T])/\mathcal B(\mathbb R)$ measurable?