The following example shows that if $D[0,1]$ (Skorokhod space of cadlag functions) is equipped with the Borel $\sigma$-field $\mathcal{B}(D[0,1])$ generated by the closed sets under the uniform metric, the empirical d.f. $G_n$ will not be a random element in $D[0,1]$, i.e., $G_n$ is not $\mathcal{A} / \mathcal{B}(D[0,1])$-measurable.
Example. Consider $n=1$ so that $G_1(t)=1_{[0, t]}(U_1), t \in[0,1]$, visualize the random function $G_1$ over $[0,1] ; G_1(t)=1_{t \geq U_1}$.
Let $B_s$ be the open ball in $D[0,1]$ with center $1_{[s, 1]}$ and radius $1 / 2$, where $s \in [0,1]$. For each subset $A \subset[0,1]$ define $$ E_A:=\cup_{s \in A} B_s . $$
Observe that $E_A$ is an open set as an uncountable union of open sets is also open.
If $G_1$ were $\mathcal{A} / \mathcal{B}(D[0,1])$-measurable, the set { $ \omega \in \Omega: G_1(\omega) \in E_A $ }= { $\omega \in \Omega: U_1(\omega) \in A $} would belong to $\mathcal{A}$.
A probability measure could be defined on the class of all subsets of $[0,1]$ by setting $\mu(A):=\mathbb{P}\left(U_1 \in A\right)$. This $\mu$ would be an extension of the uniform distribution to all subsets of $[0,1]$.
Unfortunately such an extension cannot exist! Thus, we must give up Borel measurability of $G_1$. The argument can be extended to $n \geq 1$ (see Problem 1 in [Pollard, 1984, Chapter IV]).
The above example shows that the Borel $\sigma$-field generated by the uniform metric on $D[0,1]$ contains too many sets.
Q: I do not understand fully the intuition behind the example. In particular, why the uniform metric induces a sigma field with too many sets? A good sigma field should not let $E_A$ to be measurable, right? Can anyone help with this?
Thank you very much.