# What is vague convergence and what does it accomplish?

For convenience, let's say that I have a locally compact Hausdorff space $X$ and am concerned with probability measures on its Borel $\sigma$-algebra $\mathcal{B}(X)$. Natural vector spaces to consider might be $$\mathcal{M}_b(X) \subseteq \mathcal{M}(X)$$ of $\mathbb{C}$-valued Radon measures, where $\mathcal{M}_b(X)$ are those measures with bounded total variation with norm given by $$\|\mu\|_{TV} = \sup \sum_{j} |\mu(A_j)|$$ where the supremum is taken over all countable disjoint unions $\bigcup_j A_j = X$ of Borel sets $A_j \in \mathcal{B}(X)$. Let $\mathcal{M}_b^+(X)$ and $\mathcal{M}^+(X)$ denote those Radon measures that take only values in the nonnegative real numbers. Finally, let $\mathcal{P}(X)\subseteq \mathcal{M}_b^+(X)$ denote the probability measures.

We can also define the vector spaces $$\text{C}_c(X) \subseteq \text{C}_0(X) \subseteq \text{C}_b(X) \subseteq \text{C}(X)$$ of continuous functions with compact support, almost-compact support (arbitrarily small off a compact subset), bounded. The first three can be equipped with the supremum norm, under which the closure of $\text{C}_c(X)$ is $\text{C}_0(X)$ which is in turn a closed subalgebra of $\text{C}_b(X)$.

On the other hand, $\text{C}(X)$ can be equipped with the compact-open topology, the topology of uniform convergence on compact subsets. Sometimes the space $\text{C}_c(X)$ is also equipped with this topology, so I will distinguish what topology is being used by writing $\text{C}_c^{\sup}(X)$ for the normed vector space under the supremum norm and $\text{C}_c^{co}(X)$ for the locally convex topological vector space.

According to Wikipedia, the dual of $\text{C}_0(X)$ as a Banach space with supremum norm is known to be $\mathcal{M}_b(X)$, whose norm as a Banach dual space is given by the total variation. Additionally, the positive cone in the dual of $\text{C}^{co}_c(X)$ is identified with $\mathcal{M}^+(X)$, although the topology in that article on $\text{C}_c(X)$ is not specified. It is, however, specified here.

For probability measures, we say that $\mu_n \to \mu$ weakly if we have convergence against every test function from $\text{C}_b(X)$. As is mentioned in the answer here, the notion of vague convergence in probability is convergence against every test function from either $\text{C}_c(X)$ or $\text{C}_0(X)$, although these are not the same.

First Question: It seems to me that the dual of $\text{C}_c(X)$ should be $\mathcal{M}(X)$, but the topologies of $\text{C}^{co}_c(X)$ and $\text{C}_c^{\sup}(X)$ are not the same, so which dual is it? What is the dual of the other one? EDIT: As mentioned in the comments, the dual of $\text{C}_c^{\sup}(X)$ is, as a Banach space, the same as the dual of $\text{C}_0(X)$, while the dual of $\text{C}_c^{co}(X)$ is presumably something else.

Second Question: What is the advantage in defining vague convergence with respect to $\text{C}_c(X)$ or $\text{C}_0(X)$? For instance, if $X$ is separable, then the closed unit ball in the dual of $\text{C}_0(X)$ is weak-$*$ metrizable since $\text{C}_0(X)$ is a Banach space, so the topology of vague convergence would be metrizable. I'm not positive if the same is true if we take the closed unit ball in the dual of $\text{C}_c^{\sup}(X)$ since it is only a normed space.

Third Question: If $\text{C}_c(X)$ is used to define vague convergence on probability measures, it doesn't matter for the definition if we use $\text{C}_c^{\sup}(X)$ or $\text{C}_c^{co}(X)$, but it does matter for proving theorems about the convergence. Banach–Alaoglu only applies to the dual of $\text{C}_c^{\sup}(X)$ because it is a normed vector space, and this tells us something about probability measures (namely that sequences contain vague-convergent subsequences to other measures in the unit ball). What possible motivation would there be to describe vague convergence in terms of $\text{C}_c^{co}(X)$ instead?

Fourth Question: Unless I am missing something important, the elements $\mathcal{M}_b(X)$ are also linear functionals on $\text{C}_b(X)$, although there are additional elements of its dual that are not expressible as measures. Weak convergence in probability is given in relation to functions in $\text{C}_b(X)$, so it seems this would make it weak-$*$ convergence with respect to $\text{C}_b(X)$. Is there any problem with thinking this way?

It seems like the motivation here is being able to guarantee certain types of convergence. For instance, it would be nice if when a sequence of probability measures converged to a non-degenerate measure, that measure was also a probability measure. Or it would be nice to have convergence that was metrizable. Or it would be nice to have convergence described by the pointwise convergence of some proxy functions, such as the Fourier transform for weak convergence and the Stieltjes transform for vague convergence. Or it would be nice for the space of probability measures to be compact under the topology corresponding to a type of convergence.

A suppose a broader question is: how do weak and vague topologies (and whatever topologies are induced by $\text{C}_0(X)$ weak-$*$, or dual norm topologies from $\text{C}_0(X)$ and $\text{C}_b(X)$) work to accomplish these goals, and when does one succeed while the others fail?

• $C_c^{sup}$ is dense in $C_0$, so they have the same dual. For the fourth question: No, there is no problem with thinking this way. The weak convergence in probability is the weak$\ast$ convergence (from $C_b$) in functional analysis. – user95282 Mar 13 '17 at 2:07
• Right, of course, thanks, their duals will be the same as sets and have the same Banach space topology, but the weak-* topologies from $C_0$ and $C_c^{sup}$ will be different, won't they? – Greg Zitelli Mar 13 '17 at 3:11
• $C_c(X)$ is dense in $C^{co}(X)$ and hence both spaces have the same dual which is the space of (complex) measures with compact support. – Jochen Wengenroth Mar 13 '17 at 14:50
• The vague topologies from either $C_0$ or $C_c^{\mathrm{sup}}$ induce the same topology on the set of all subprobability measures (positive measures with total measure $\le 1$), and this topology is compact. That is a major reason for using the vague topology. – Nate Eldredge Mar 14 '17 at 4:19