For quite a long time, I have been confused about the definitions of weak convergence and vague convergence of measures among other modes of convergence that root from functional analysis, mainly due to many different definitions and theorems from probability books. I would appreiciate it if someone can clarify the terms and give a clear picture of the concepts. (Note that Did has answered some of my related questions before. Thank you, Did!)

  1. In Kallenberg's probability book, he defines weak convergence of a sequence measures to be

    Consider any probability measures $\mu$ and $\mu_1, \mu_2, \dots$ on some metric space $(S, \rho)$ with Borel a-field $S$, and say that $\mu_n$ converges weakly to $\mu$, if $\int f d\mu_n \to \int f d\mu$ for every $f \in C_b(S)$, the class of bounded, continuous functions $f: S \to \mathbb R$.

  2. Kallenberg defines vague convergence of a sequence of measures to be

    Consider the space $\mathcal M = \mathcal M(\mathbb R^d) $of locally finite mea- sures on $\mathbb R^d$. On $\mathcal M$ we may introduce the vague topology, generated by the mappings $\mu \mapsto \int f d\mu$ for all $f \in C_K^+$, the class of continuous functions $f: \mathbb R^d \to \mathbb R_+$ with compact support. In particular, $\mu_n$ is said to converge vaguely to $\mu$ if $\mu_n f \to \mu f$ for all $f \in C_K^+$. If the $\mu_n$ are probability measures, then clearly $\mu(\mathbb R^d) < 1$.

  3. Folland in his real ananlysis book defines vague topology and therefore vague convergence for complex Radon measures on a locally compact Hausdorff (LCH) space $X$ as weak* topology and weak* convergence wrt $C_0(X)$. He says the term "vague" is common in probability theory, and has the advantage of forming an adverb more gracefully than "weak*". The vague topology is sometimes called the weak topology, but this terminology conflicts with his, since $C_0(X)$ is rarely reflexive.
  4. In Kai Lai Chung's probability book, a sequence of subprobability measures $\mu_n$ on $\mathbb R$ are defined to vaguely converge to another subprobability measure $\mu$, if there exists a dense subset $D$ of $\mathbb R$ s.t. $\forall a, b \in D, a < b, \mu_n((a,b]) \to \mu((a,b])$.
  5. Next in Chung's, Theorem 4.4.1 says in case of subprobability measures, vague convergence is equivalent to weak* convergence wrt $C_0(\mathbb R)$ and $C_K(\mathbb R)$. Theorem 4.4.2 says in case of probability measures, vague convergence is equivalent to weak* convergence wrt $C_b(\mathbb R)$.

I was wondering if the above definitions of weak convergence and vague convergence are all weak* convergence, in the sense that the measures form (a subset of) the continuous dual of $C_b$, $C_K$, $C_K^+$, and $C_0$?

When defining vague convergence and vague topology, why does kallenberg use $C_K^+$ instead of $C_K$, Folland use $C_0$, and Kai Lai Chung uses $C_K$, $C_0$ and $C_b$? Are their definitions of vague convergence consistent with each other?

Among the convergences of measures wrt $C_b$, $C_K$, $C_K^+$, and $C_0$, when does which imply which? When is which equivalent to which?

The last question is to see if there are some unifications of the above concepts. Can the above definitions be generalized to more general measures (probability measures, subprobability measures, locally finite measures are used in the definitions above), and to more general underlying spaces (metric space, $\mathbb R^d$ and $\mathbb R$ are used in the definitions above)?

Thanks and regards!!

  • 2
    $\begingroup$ There are a lot of questions here and I don't think I can answer them all, so I'll just leave this part as a comment. A critical fact for probability is that vague convergence of subprobability measures is what you get with $C_0$ or $C_K$ test functions (that is, compactly supported or vanishing at infinity) while weak convergence, which is what you get with $C_b$ test functions, is equivalent to vague convergence plus tightness. Intuitively, compactly supported functions can't tell if mass escapes to infinity, while bounded functions can. Example: $\delta_n \to 0$ vaguely, but not weakly. $\endgroup$ Commented Feb 25, 2013 at 17:19
  • $\begingroup$ @ChrisJanjigian: Thanks! I am happy to learn that! I know there are many, but I think putting them together may help to reveal a clearer picture. $\endgroup$
    – Tim
    Commented Feb 25, 2013 at 17:36
  • $\begingroup$ @ChrisJanjigian: (1) If I understand your comment correctly, for subprobability measures, weak convergence is equivalent to vague convergence plus tightness? is the measure space assumed to satisfy some other conditions? What is some reference for the statement? (2) Can the measures be generalized to be more general than subprobability measures? (3) If the subprobability measures happen to be probability measures, do we still need tightness for the equivalence? $\endgroup$
    – Tim
    Commented Feb 27, 2013 at 17:15
  • $\begingroup$ (4) "A critical fact for probability is that vague convergence of subprobability measures is what you get with C0 or CK test functions". Are convergences wrt C0 and with CK still equivalent for more general measures than subprobability measures? $\endgroup$
    – Tim
    Commented Feb 27, 2013 at 17:26
  • $\begingroup$ (1) actually I assumed that we are working on a Polish space when I said that. That is the usual assumption for the equivalence of weak convergence and vague convergence + tightness. You can find a proof of that result in Billingsley's Convergence of Probability Measures as theorem 5.2. (2) The condition of tightness is fairly specific to finite measures and you can always assume a finite measure with positive mass is a probability measure by rescaling, so I am not sure what you mean. (3) The same counterexample as above shows that you need tightness. $\endgroup$ Commented Feb 27, 2013 at 17:53

1 Answer 1


Ok so there is a lot to say here, let's start with the easiest question.

The use of $C^+_K$ instead of $C_K$ change absolutely nothing because we can always write $f=f^+-f^-$ with $f^+=\max(f,0)$ and $f^-=\max(-f,0)$ which are positive and integrable.

In order to avoid confusions between the different names of the different convergence i will talk about $C_K$, $C_0$ or $C_b$ convergence.

The simple case

First let's see what happen in the nicest case : Take $X$ to be a compact Hausdorff set. Then $C_K(X)=C_b(X)=C^0(X)$ with $C^0(X)$ the space of all continuous functions from $X$ to $\mathbb R$ (or $\mathbb C$, but it doesn't really matter). It's a bit difficult to say what $C_0(X)$ is for $X$ a topological space with no other structure but since here $X$ is compact the only natural definition is $C_0(X)=C_b(X)$. Moreover, using the Riesz-Markov theorem we can identify the (topological) dual of $C^0(X)$ and the space $\mathcal M (X)$ of all (signed and finite) Radon measures on $X$. So now we can define the weak-* topology on $\mathcal M (X)$ by $\mu_n\to \mu$ if and only if $\int fd\mu_n \to \int f\mu$ for every $f\in C^0(X)$. Obviously all the definitions you have seen before are the same here because $C_K(X)=C_0(X)=C_b(X)=C^0(X)$. So in this case all those different convergences are the same and are, in fact, the weak-* convergence of measures. The weak-* topology of a Banach has a very important propety : the closed unit ball for the strong topology is compact for the weak-* topology (it's the Banach alaoglu theorem). Which imply the following : given a bounded sequence of Radon measures $\mu_n$ there exists a Radon measure $\mu$ and a subsequence $\mu_{n_k}$ such that $\mu_{n_k}$ is weak-* convergent to $\mu$. Moreover we can prove that if the $\mu_n$ are probability measures then $\mu$ is also a probability measure (we'll see that later).

Now in the more general case when $X$ is Hausdorff but only localy compact it's more difficult. First of all the spaces $C_K$, $C_b$ and $C^0$ are all different. And when $C_0(X)$ is easily definable (when $X$ is a metric space or a topological vector space for example) it's also different from all the previous spaces. So the chances are that the different definitions of convergences won't coincide anymore. First let's see what are the differences with the previous case.

The things that still work, or not

If we look at the Riesz Markov theorem it still work with $X$ only being locally compact, but the Radon measure are not finite anymore, only locally finite. So the convergence using $C_K(X)$ functions is still the weak-* convergence of measures. The Banach Alaoglu theorem also still work but the property about probability measure doesn't hold anymore : in $\mathbb R$ the sequence $\delta_n$ is weak-* convergent to $O$, which is not a probability measure. However if the sequence $\mu_n$ is bounded and converge to $\mu$ then $\mu$ is also bounded and $\mu(X)\leq \lim\mu_n(X)$ (with the possibility of a strict inequality, as in the example before).

For the convergence with $C_0(X)$ functions we also have a representation theorem for the dual of $C_0(X)$ in terms of measures (see the wikipedia page of the Riesz-Markov theorem). So what i've said for $C_K(X)$ convergence will still hold. However even if the $C_K$ and the $C_0$ convergences can be viewed as weak-* convergences they are not the same. Take the sequence $n\delta_n$, it's $C_K$ convergent to $0$ but not $C_0$ convergent (take for example $X=\mathbb R$ and $f(x)=sin(x)/x$).

For the convergence with $C_b(X)$ functions it's different. There exists elements of the dual of $C_b(X)$ which cannot be represented as measures on $X$, for an exemple of that you can look at "Banach limit", the Banach limit of the $l^\infty$ sequence $(f(n))_{n\in\mathbb N}$ is one of those weird element of the dual of $C_b(X)$. So the $C_b$ convergence is not a weak-* convergence.

Last thing to say : since $C_K(X)\subset C_0(X) \subset C_b(X)$ we immediatly see that the $C_b$ convergence imply $C_0$ convergence which in turn imply $C_K$ convergence but the converses are all false. For example $\delta_n$ in $\mathbb R$ is $C_0$ convergent to $0$ but not $C_b$ convergent (take the bounded function sinus for example).

How to get things to work again

The almost obvious answer is : we have to work on a compact again. The usefull notion here is the notion of tight sequence (especially when dealing with probability): $(\mu_n)$ is called tight if, for all $\varepsilon>0$ there exists a compact $K_\varepsilon$ such that $\mu_n(K^C_\varepsilon)<\varepsilon$ for all $n$. So the sequence of measures are all almost supported in a compact set, so there is no possibility of mass "escaping at infinity" as it was the case with $(\delta_n)$. If $(\mu_n)$ is tight then the $\mu_n$ are finite (we assume that all the measures are locally finite)

Now you can read this answer of mine about tight sequences : Defining weak* convergence of measures using compactly supported continuous functions

To sum up things : if your bounded sequence of measures is tight then the 3 definitions are equivalent. Moreover if the $\mu_n$ are probability measure then so is the limit, and if you have a sequence of probability measure converging to a probability measure $\mu$ in the $C_K$ definition then it is tight. Moreover if $(\mu_n)$ is $C_b$ convergent then is is automatically tight.

The other questions

Locally finite measure that are note finite are usually not in the dual of $C_b$ and $C_0$. So the only definition of convergence that make sense here is the $C_K$ convergence. The definition 4 is consistent with the other definitions of vague convergence because the linear combinations of $\mathbf 1_{[a;b[}$ are dense in $C_0$ and because the sequence $(\mu_n)$ is bounded.

If you read french you can learn more on http://www.proba.jussieu.fr/cours/dea/telehtml/telehtml.html

  • 3
    $\begingroup$ Just a little comment concerning the "very important propety": The weak*-topology does NOT make Banach space locally compact. The Banach alaoglu theorem tells you that the unit ball is compact w.r.t. the weak*-topology but it fails to be zero-neighbourhood with that topology. While with the norm topology, the unit ball is a neighbourhood but not compact. (Assuming all spaces are infinite dimensional). Generally speaking: An infinite dimensional real topological vector space is NEVER locally compact. But as I see it, the rest of the answer should still be correct. $\endgroup$
    – Tom
    Commented Oct 12, 2018 at 8:45
  • 2
    $\begingroup$ Thanks for the comment, you are obviously right. I wanted to say that every point has a strong neighborhood that is compact for the weak-* topology, but it came off wrong. I corrected my mistake. I should point out an other mistake : the $C_b$ convergence is in fact a weak-* convergence since measures are in the dual of $C_b$. The only difference is that there are more elements in that dual than just the (signed) Radon measure. I should also edit that one day. $\endgroup$
    – Renart
    Commented Oct 12, 2018 at 14:16
  • $\begingroup$ @Renart Thank you so much for your time to explain these definitions but I still have a question if you could look at it: If the space X is not locally finite then what would be the definition of vague convergence ? In particular, I want to know what would be the space $\mathcal{M} (X)$ in that case? $\endgroup$ Commented Jan 7, 2021 at 22:13

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