# Weak convergence of probability measures, Borel measurability and separable spaces

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?

• 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 $any$ $H\subset[0,1]$. – Jack London Oct 27 '17 at 12:23
• 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 $X_n^{-1}( \cup_{i \geq 1} S_i) = \cup_{i\geq 1} X_n^{-1}(S_i)$. – Jack London Oct 27 '17 at 12:39
• 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 $(\mathbb{D}, d)$. – Jack London Oct 27 '17 at 12:44
• Sorry, above I should have said: the cardinality of the power set of $[0,1]$ is larger than the cardinality of $\mathscr{B}$. – Jack London Oct 27 '17 at 12:49
• 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. – Jack London Oct 27 '17 at 13:34