Denote by $D(0,T)$ the space of right continuous functions with left limits defined on $[0,T]$. Let $\mathbb P$ be a probability measure on $D(0,T)$. Define
$$cont(\mathbb P):=\Big\{t\in [0,T]:~ \mathbb P\big(\big\{f\in D(0,T):~ f(t-)=f(t)\big\}\big) = 1\Big\}.$$
Is $cont(\mathbb P)$ dense in $[0,T]$?
I strongly believe the answer is yes, but cannot find a reference.
Let $f\in D(0,T)$. Let $\mathsf{Disc}_n(f)=\{t\colon |f(t^+)-f(t^-)|\ge \frac 1n\}$. I claim this set is discrete. If not, there is a sequence of distinct points $(t_k)\in\mathsf{Disc}_n(f)$ converging to some $t_0$. There is then a subsequence converging to $t_0$ consisting entirely of points on the left or entirely of points on the right. This contradicts the left limit or right continuity property at $t_0$. Hence $\mathsf{Disc}_n(f)$ is discrete, and therefore finite. Hence $\mathsf{Disc}(f)=\bigcup_n\mathsf{Disc}_n(f)$ is countable and so of measure 0.
Now consider $$ S=\{(f,t)\colon f\in D(0,T),\, t\in\mathsf{Disc}(f)\}. $$ By the above calculation, for each $f\in D(0,T)$, $\text{Leb}(\mathsf{Disc}(f))=0$. Hence $\mathbb P\times\text{Leb}(S)=0$. So by Fubini's theorem, $$ \mathbb P\times\text{Leb}(S)=\int_0^T \mathbb P(\{f\in D(0,T)\colon t\in\mathsf{Disc}(f)\})\,dt=0. $$ Hence for $\text{Leb}$-a.e. $t$, $$ \mathbb P(\{f\in D(0,T)\colon t\in\mathsf{Disc}(f)\})=0.$$
