For a topological space $E$ denote $\mathscr{B}(E)$ its Borel $\sigma$-algebra. For a measurable space $(A, \mathscr{A})$ and $B \subseteq A$ denote $\mathscr{A}|B$ as the trace $\sigma$-algebra on $B$.

Let $E$ be a Polish space (i.e. separable and completely metrizable) and $\mathscr{E} := \mathscr{B}(E)$ its Borel $\sigma$-algebra. Consider the measurable product space $E^{[0, 1]}$ equipped with the product $\sigma$-algebra $\mathscr{E}^{\otimes [0,1]}$.

  1. The space $C := C([0, 1], E)$ of continuous functions with the topology given by the uniform convergence is Polish. It is a subset of $E^{[0,1]}$ but is not contained in $\mathscr{E}^{\otimes [0,1]}$. However, $\mathscr{E}^{\otimes [0,1]} | C = \mathscr{B}(C)$.

  2. The Skorokhod space $D := D([0, 1], E)$ of cadlag functions is also Polish. It is a subset of $E^{[0,1]}$ but is also not contained in $\mathscr{E}^{\otimes [0,1]}$. Here, it also holds that $\mathscr{E}^{\otimes [0,1]} | D = \mathscr{B}(D)$. Moreover, the relative topology of $D$ onto $C \subseteq D$ coincides with the topology on $C$ given by the uniform convergence.

In this way, one can say that the space $D$ "extends" the space $C$ in $E^{[0,1]}$. This extension is an extension considered as Polish spaces and as such that their trace $\sigma$-algebras coincide with their Borel $\sigma$-algebras.

Question: Can the space $D$ further be extended in this meaning? If yes, is there some canonical meaning of a "maximal" extension, i.e. a subset $X \subseteq E^{[0,1]}$ that can be given a topology under which

  1. $X$ is Polish
  2. $\mathscr{E}^{\otimes [0,1]} | X = \mathscr{B}(X)$
  3. for every Polish subspace $Y \subseteq X$ the relative topology from $X$ to $Y$ coincides with the topology on $Y$?

Of course, there are also other spaces that extend $C$, e.g. the Polish space $\widetilde{D}$ of left-continuous right-limit functions and it holds that $C = D \cap \widetilde{D}$.


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