Let
$\mathcal{D}:=\{\pi_{t_1,\ldots, t_d}^{-1}(B) : {t_1,\ldots, t_d} \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R}^d)\}$ and
$\mathcal{D}_0:=\{\pi_{t}^{-1}(B) : t\in [0,1]$ and $B \in \mathcal{B}(\mathbb{R})\}$,
where $\pi_t: (D[0,1],\mathcal{B}_{(D[0,1],d_S)})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ and $\pi_{t_1,\ldots, t_d}$ the same for $\mathbb{R}^d$. $d_S$ denotes the Skorohod-metric making $(D[0,1],d_S)$ into a separable and complete space. My questions:
In the literature it is always mentioned (without proof) that $\sigma(\mathcal{D})=\sigma(\mathcal{D}_0)$, how can this be proved?
I want to prove that $\sigma(\mathcal{D})=\mathcal{B}_{(D[0,1],d_S)})$. Most of the time the authors refer for the proof to Billingsley. Unfortunately, his proof is quite hard to read and without many details. Does anyone know a referene (or proof) which is easier to handle?
One could also look at the space equipped with uniform norm and Borel $\sigma$-field $\mathcal{B}_{(D[0,1],d_U)}$. It follows from the properties of the metrics that $\mathcal{B}_{(D[0,1],d_S)}\subset\mathcal{B}_{(D[0,1],d_U)}$ and therefore (from 2.) $\sigma(\mathcal{D})\subset\mathcal{B}_{(D[0,1],d_U)}$. Can anyone give a set which belongs to $\mathcal{B}_{(D[0,1],d_U)}$, but not to $\sigma(\mathcal{D})$?