Most of the following comes from studying the first pages of van der Vaart and Wellner (1996). Let $(\mathbb{D},d)$ be a metric space and let $\{P_n, \, n\geq 1\}$ and $P$ be Borel probability measures on $(\mathbb{D},\mathscr{D})$, where $\mathscr{D}$ is the Borel $\sigma$-field on $\mathbb{D}$. Then we say that $P_n$ converges weakly to $P$ if and only if

$\int_\mathbb{D} f \, dP_n \to \int_\mathbb{D} f \, dP, \quad$ for all $f \in C_b(\mathbb{D})$.

Equivalenty, letting $X_n$ and $X$ be $\mathbb{D}$-valued random elements with distributions $P_n$ and $P$ respectively, then $X_n$ converges in distribution to $X$ if and only if

$\mathbb{E}f(X_n)\to \mathbb{E}f(X), \quad$ for all $f \in C_b(\mathbb{D})$.

Letting $(\Omega_n, \mathscr{A}_n)$ be the underlying measure spaces on which the maps $X_n$ are defined, what we are assuming is that $X_n^{-1}(D)\in \mathscr{A}_n$, for every Borel set $D \in \mathscr{D}$. This required measurability usually holds when $\mathbb{D}$ is a separable metric space (e.g. $\mathbb{R}^k$ or $[0,1]$ endowed with the sup-norm).

Now, one is often interested into the case in which $X_n$ is a random element of

$\bullet$ $(\mathbb{D},d)=(\ell_\infty(\mathcal{F}),\Vert \cdot \Vert_\mathcal{F})$, the space of bounded functions on a given space $\mathscr{F}$ endowed with the sup-norm;

$\bullet$ $(\mathbb{D},d)=(D[0,1],\Vert \cdot \Vert_{\infty})$, the Scohorod space of càdlàg functions on the closed unit interval endowed with the sup-norm.

These spaces are not separable.

Let's consider an example: $\xi_1,...,\xi_n$ are i.i.d. according to the uniform distribution on $[0,1]$ and defined as coordinate projections on the product probability space $([0,1]^n, \mathcal{B}^n, \lambda^n)$ where $\mathcal{B}$ and $\lambda$ denote the Borel $\sigma$-field and the Lebesgue measure on the unit interval. Letting

$X_n=\left(\, n^{-1}\sum_{i=1}^n 1_{[0,t]}(\xi_i) \, \right)_{t \in [0,1]}$,

we have that the inclusion $X_n^{-1}(\mathscr{D}) \subset \mathcal{B}^n$ fails to hold, since the Borel $\sigma$-field induced by the topology of the sup-norm $\mathscr{D}$ is now "too large". Why is it so? Which is the actual link between separability and Borel-measurability? Moreover, the concept of weak convergence as in the first two displays above can not be employed to study the weak limit of $X_n$, beacause of the lack of Borel-measurability. Intuitively, why the classical theory of weak convergence has been constructed relying on this "specific" measurability requirement?

  • $\begingroup$ In order to answer to a part of the question, I report a summary from pp 150-153 in Billingsley (1968). Let for simplicity n=1. Then in the example $X_n(t)=X_1(t)=1_{[0,t]}(\xi_1)$. Denoting by $A_\theta$ a sphere with center $1_{[0,\theta]}$ and radius $1/2$, we have that for any $H\subset [0,1]$, $\cup_{\theta \in H}A_\theta \in \mathscr{D}$, since it is open (observe that the lack of separability does not allow us to express such set as a countable union of spheres). Moreover, observe that $X_1^{-1}(\cup_{\theta \in H}A_\theta)=\{\omega:\xi_1(\omega) \in H\}$, for $any$ $H\subset[0,1]$. $\endgroup$ – Jack London Oct 27 '17 at 12:23
  • $\begingroup$ If $X_1$ was Borel measurable, we would have that $\{\omega: \xi_1(\omega) \in H \} \in \mathscr{B}$, for $any$ $H\subset [0,1]$. This is not possible, since the power set of $[0,1]$ strictly contains $\mathscr{B}$. This explains why $X_1^{-1}(\mathscr{D})\subset\mathscr{B}$ can't be true. Intuitively speaking, without separability of $(\mathbb{D},d)$, the class of open sets is "too reach". If $(\mathbb{D},d)$ is separable, every open set $D$ would be in the form $\cup_{n\geq1}S_i$, where $S_i$ are spheres, and clearly $X_n^{-1}( \cup_{i \geq 1} S_i) = \cup_{i\geq 1} X_n^{-1}(S_i)$. $\endgroup$ – Jack London Oct 27 '17 at 12:39
  • $\begingroup$ Therefore, in the latter case, Borel measurability is guaranteed by measurability of sets in the form $X_n^{-1}(S)$, where $S$ is a sphere in $(\mathbb{D}, d)$. $\endgroup$ – Jack London Oct 27 '17 at 12:44
  • $\begingroup$ Sorry, above I should have said: the cardinality of the power set of $[0,1]$ is larger than the cardinality of $\mathscr{B}$. $\endgroup$ – Jack London Oct 27 '17 at 12:49
  • $\begingroup$ Final remark: in probability we are able to "construct" measures on cylindrical $\sigma$-fields (in Banach spaces, the coarsest $\sigma$-fields s.t. every continuous linear function is measurable). A cylinder set is the natural open set of a product topology (space of marginal distribution) and cylinder set measure (or promeasure, or premeasure, or quasi-measure, or CSM) is a kind of prototype for a measure on an infinite-dimensional vector space. Now, unlike in the finite dimensional case (e.g. $\mathbb{D}=\mathbb{R}^k$) cylindrical and Borel $\sigma$-field do not usually coincide. $\endgroup$ – Jack London Oct 27 '17 at 13:34

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