On the Skorohod space $D[0,1]$ of cadlag functions one usually considers either the uniform norm $\|\cdot\|_{\infty}$ or the $J_1$-metric $\varrho$. I was wondering whether both generate the same Borel $\sigma$-algebra:
It is widely known that
$$\varrho(f,g) \leq \|f-g\|_{\infty},$$
i.e. the $J_1$-topology is coarser, and this implies in particular that the corresponding Borel $\sigma$-algebras satisfy the relation
$$\mathcal{B}((D[0,1],\varrho)) \subseteq \mathcal{B}((D[0,1],\|\cdot\|_{\infty}). \tag{1}$$
On the other hand, one can show that $\mathcal{B}((D[0,1],\varrho))=\sigma(\pi_t;t \in [0,1])$ where $\pi_t(f) :=f(t)$, $t \in [0,1]$, denotes the natural projection (see e.g. Billingsley, Convergence of Probability Measures, Theorem 12.5). Now for any countable dense set $T \subseteq [0,1]$, $1 \in T$, we have
$$B_{(D[0,1],\|\cdot\|_{\infty})}[0,1] := \{f \in D[0,1], \|f\|_{\infty} \leq 1\} = \bigcap_{t \in T} \{f \in D[0,1]; \pi_t f \in [-1,1]\} \in \sigma(\pi_t;t \in [0,1]) \stackrel{(1)}{=} \mathcal{B}((D[0,1],\varrho)).$$
by the right-continuity of the functions. A similar argumentation shows that $B_{(D[0,1],\|\cdot\|_{\infty})}[f,\varepsilon] \in \mathcal{B}((D[0,1],\varrho))$ for any $\varepsilon>0$, $f \in D[0,1]$. Hence,
$$\mathcal{B}((D[0,1],\varrho)) \supseteq \mathcal{B}((D[0,1],\|\cdot\|_{\infty}),$$
and this means that both Borel-$\sigma$-algebras are equal. Is there anything wrong about this argumentation?
Best Answer
You show that the ball-sigma-algebra of the uniform topology coincides with the Skorohod-Borel-sigma-algebra. However, the Borel-sigma-algebra of the uniform topology is strictly larger (that can happen because the uniform topology is not separable here). See chapt. 15 of Billingsley's book for more on this.