Edit: Changed from "Hausdorff" to "metric" spaces.

Let $\mathcal{M}(\Omega)$ denote the space of signed regular Borel measures on a compact metric space $\Omega$. By Riesz-Markov, this is the dual space of $C(\Omega)$, the space of all continuous real valued functions on $\Omega$. Denote by $$\mathcal{P}(\Omega) = \{\mu\in\mathcal{M}(\Omega)\ :\ \mu\geq 0, \mu(\Omega)=1\}$$ i.e. the set of all probability measures in $\mathcal{M}$.

The weak convergence (also called weak* convergence) in $\mathcal{M}(\Omega)$ is defined by duality and it is known that weak convergence in $\mathcal{P}(\Omega)$ can be metrizised by, e.g. the Prokhorov metric $d_P$ or the Wasserstein metrics $d_W$.

Obviously, both metrics do not metrizise weak convergence on $\mathcal{M}(\Omega)$: For the Wasserstein metric we have $d_W(\mu,\nu)=\infty$ if $\mu(\Omega)\neq\nu(\Omega)$ and for the Prokhorov metric we do not even have $d_P(\mu,\mu)=0$, as far as I see.

Googling and searching MSC did not produce any results on my question:

Are there any metrics available which metrizise weak convergence of signed regular Borel measures?

I would be surprised if there weren't (or are there any fundamental obstructions?).

I would also be happy with metrics for weak convergence of non-negative measures (but not normed ones) or for uniformly bounded measures and would also like to know the answer to the same question for vector valued Radon measures on a metric space.

2dn edit: Thanks for the great answers. I had forgotten the general procedure to define a metric for weak(*) convergence on bounded set, but in fact I had a more "geometric" metric in mind, something in the direction of R Ws and Dans answers.

  • 3
    $\begingroup$ Do you perhaps want $\Omega$ to be metric? It is not clear to me how the Prokhorov and Wasserstein metrics on $\mathcal{P}(\Omega)$ might be defined in the absence of a metric on $\Omega$, and I am suspicious of the suggestion that $\mathcal{P}(\Omega)$ is metrisable even when $\Omega$ is not (for example in the case where $\Omega$ is the Stone-Čech compactification of $\mathbb{N}$). In fact, can we not identify $\Omega$ with the subset of $\mathcal{P}(\Omega)$ which comprises the Dirac measures, and deduce that if $\mathcal{P}(\Omega)$ is metrisable then $\Omega$ must be metrisable also? $\endgroup$
    – Ian Morris
    Jan 30, 2013 at 10:11
  • $\begingroup$ Indeed, I wanted metric spaces - question edited. Thanks for pointing this out. $\endgroup$
    – Dirk
    Jan 30, 2013 at 16:48
  • $\begingroup$ Thanks a lot for the enlightening answers so far! Since I am off for a long weekend, I will have a closer look in a few days. $\endgroup$
    – Dirk
    Jan 30, 2013 at 16:49

6 Answers 6


Of course, there are many ways of metrizing the weak topology on $\mathcal M(\Omega)$ by using various tools of functional analysis. However, as it has already been pointed out by Dan, the most natural way is to use the transportation metric on the space of measures. [It is much more natural than the Prokhorov metric. I don't want to go into historical details here - they can be easily found elsewhere, but I insist that the transportation metric should really be related with the names of Kantorovich (in the first place) and his collaborator Rubinshtein]. Dan gives its dual definition in terms of Lipschitz functions, however its "transport definition" is actually more appropriate here. Let me remind it.

Given two probability measures $\mu_1,\mu_2$ on $\Omega$ $$ \overline d(\mu_1,\mu_2) = inf_M \int d(x_1,x_2) dM(x_1,x_2) \;, $$ where $d$ is the original metric on $\Omega$, and the infimum (which is in fact attained) is taken over all probability measures $M$ on $\Omega\times\Omega$ whose marginals ($\equiv$ coordinate projections) are $\mu_1$ and $\mu_2$. One should think about such measures as "transportation plans" between distributions $\mu_1$ and $\mu_2$, while the integral in the RHS of the definition is the "cost" of the plan $M$.

It is obvious that the above definition makes sense not just for probability measures, but for any two positive measures $\mu_1,\mu_2$ with the same mass. Moreover, $\overline d(\mu_1,\mu_2)$ actually depends on the difference $\mu_1-\mu_2$ only, so that one can think about it as a "weak norm" $$ |||\mu_1-\mu_2||| = \overline d(\mu_1,\mu_2) $$ of the signed measure $\mu_1-\mu_2$ (clearly, it is homogeneous with respect to multiplication by scalars).

Let now $\mu=\mu_1-\mu_2$ be an arbitrary signed measure, where $\mu_1,\mu_2$ are the components of its Hahn decomposition. The only reason why the definition of the weak norm does not work in this situation is that the measure $\mu$ need not to be "balanced" in the sense that the total masses $\|\mu_1\|$ and $\|\mu_2\|$ need not be the same any more. However, this can be easily repaired in the following way: extend the original space $\Omega$ to a new metric space $\Omega'$ by adding to it an "ideal point" $o$ and putting $d(\omega,o)=1$ for any $\omega\in\Omega$. Then the measure $$ \mu'=\mu - (\|\mu_1\|-\|\mu_2\|)\delta_o \;, $$ where $\delta_o$ is the unit mass at the point $o$, is now balanced, so that $|||\mu'|||$ is well defined. Therefore, one can extend the definition of the weak norm $|||\cdot|||$ to arbitrary signed measures $\mu$ by putting $$ |||\mu|||=|||\mu'||| \;. $$

It is now easy to see that the distance $|||\mu_1-\mu_2|||$, where $\mu_1,\mu_2$ are two arbitrary signed measures, metrizes the weak topology on $\mathcal M(\Omega)$.

  • $\begingroup$ That is in the direction I was thinking. I just found a similar construction in Gromov's "Metric structures" chapter $3\tfrac{1}{2}$.B: For any metric $d$ defined for measures of the same total mass one may define a metric for all finite measures as follows. The distance between two measures $\mu$ and $\nu$ with total masses $\mu(\Omega) = m$ and $\nu(\Omega)=n$ with $n>m$ define $D(\mu,\nu) = n-m + d(\mu,\tfrac{m}{n}\nu)$ (and the distance to infinite measures is just $\infty$). Apparently, this definition also works for metric measure spaces. $\endgroup$
    – Dirk
    Feb 6, 2013 at 8:14
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    $\begingroup$ This is a very late comment, but how does this answer not contradict the fact that the dual space of a Banach space $X$ is weak*-metrizable if and only if $X$ is finite dimensional, as mentioned by Jochen below? It is true that the space of positive measures on $\Omega$ is metrizable, as in Theorem 3.1 of jstor.org/discover/10.2307/25048364. $\endgroup$ Mar 11, 2015 at 0:10
  • $\begingroup$ Sorry for the late answer: This thing is, that I only want to metrize the notion of weak$*$ convergence not the whole weak* topology. These are in fact two different things. The weak* topology can be metrized on bounded subsets (see Pietro Majers answer). $\endgroup$
    – Dirk
    Nov 17, 2015 at 14:10
  • $\begingroup$ As a matter of fact it sticked to the advice and using Kantorovich and Rubinstein for these norms (rather than Wasserstein or Monge), see the paper Imaging with Kantorovich-Rubinstein discrepancy. $\endgroup$
    – Dirk
    Nov 17, 2015 at 14:13
  • 1
    $\begingroup$ 10 years to late, but the extension $d(\omega,o)=1$ need not be a metric. For example, if there are points $x,y\in\Omega$ with $d(x,y)>2$, the extenion does not satisfy the triangle inequality. $\endgroup$ Jan 8, 2023 at 12:09

If $X$ is an infinite dimensional separable Banach space (like $C^0(\Omega)$, for a compact metric space $\Omega$ ), and $\{y _ j \} _ {j\ge 1}$ is a dense sequence in its unit ball, one considers the norm on $X^*$ defined by $$ ||| u |||:=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u, y _ j \rangle |\, ,$$ which is weaker than the dual norm $\| \cdot \|$, since $ |||u|||\le \| u\|$. On the space $ X ^ * $, the $|||\cdot|||$ norm topology and the $w^*$ topology differ, because the latter is not metrizable. However, they induce the same topology on any (dual norm) bounded subset.

rmk. If $Y$ denotes the dense linear subspace spanned by $\{y _ j \} _ {j\ge 1}$ we may also consider the weak topology $\sigma(X ^ *,Y)$ on $ X ^ * $: that is, the smallest TVS topology that makes continuous any evaluation map at points $y\in Y$, that is $ X ^ * \ni u\mapsto \langle u, y\rangle $. With this topology, $X ^ * $ is a locally convex, Hausdorff and first countable space, though metrizable, yet not normable as not locally bounded. So this is strictly weaker than the above $|||\cdot|||$ norm-topology; it is also strictly weaker than the weak $^*$ topology $\sigma(X ^ *,X)$, since their dual spaces are the evaluations at points of $Y$, respectively of $X$. Again, on $\|\cdot\|$-bounded subsets they induce the same topology: indeed, if $u_n\to u$ in $\sigma(X ^ *,Y)$ with $\|u_n\|\le R$, then $ ||| u - u _ n |||=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u - u_n , y _ j \rangle |$ also converges to $0$, since the terms of the series converge to $0$ while dominated by the series $ \sum _ {j=1} ^\infty 2 ^ {-j} (2R) $. Moreover, it is also true that $u_n\to u$ in $\sigma(X ^ *,X)$, because for any $x\in X$ and $y\in Y$ $|\langle u - u_n , x \rangle |\le |\langle u - u_n , y \rangle | + 2R\| x - y \|$, whence $\limsup _ { n \to \infty} |\langle u - u _ n , x \rangle | \le 2R\| x - y \| $, and the RHS can be made arbitrarily small.

  • $\begingroup$ Nice answer. It is worth to point out that if $u_j$ is unbounded (with respect to $||\cdot ||$), then it might still be the case that $|||u_n|||\to 0$ but of course $u_j$ doesn't converge weakly. To "see" why this is, notice that $u_j(y_j)\to 0$ for all $y_j$ but not uniformly in $j$. For a generic $y$ not equal to any $y_j$ we can only bound $|u_n(y)|\leq |u_n(y-y_j)|+|u_n(y_j)|\leq ||u_n|| |y_j-y| +|u_n(y_j)|$. So, for a given $n$, if we try to take $j$ so that $||u_n|| |y_j-y|$ is small, it might be the case that $||u_n(y_j)||$ gets large, and $|u_n(y)|$ doesn't go to $0$. $\endgroup$
    – No-one
    Nov 29, 2023 at 0:22
  • $\begingroup$ Of course this is a hand-waving argument just to build some intuition, a proper proof follows from the fact that every weak convergence sequence is (strongly) bounded. $\endgroup$
    – No-one
    Nov 29, 2023 at 0:41
  • $\begingroup$ Moreover, in the specific case of positive Radon measures on a compact $K$ it is true that $|||\cdot|||$ metrizes the weak$^*$ topology. Indeed one can just take $y_0\in C(K)$ with $y_0(x)=1$ for all $x$, and then the convergence in $|||\cdot|||$ implies that the masses of the $u_n$ are bounded. This is what is done in Theorem $3.1$ of the paper jstor.org/stable/25048364 linked by Iian Smith. $\endgroup$
    – No-one
    Nov 29, 2023 at 0:48

In the case that $\Omega$ is a metric space, the Kantorovich-Rubinstein norm (or the equivalent Bounded Lipschitz norm) will meet some of your needs. It will metrize weak convergence for nonnegative measures, but not signed measures. It has the advantage, however, of being particularly pleasant to work with. It is defined by

$\|\mu\| := \sup\left\{\int f \,d\mu : f \in Lip_1(\Omega), \ \sup_{\omega}|f(\omega)| \le 1\right\}$,

where $Lip_1(\Omega)$ is the set of 1-Lipschitz functions from $\Omega$ to $\mathbb{R}$. This defines a norm on $\mathcal{M}(\Omega)$, and the metric $d(\mu,\nu) := \|\mu - \nu\|$ metrizes weak convergence on the space $\mathcal{M}^+(\Omega)$ when $\Omega$ is separable. By Kantorovich duality, $d$ coincides with a Wasserstein metric (generated by the metric of $\Omega$ capped by 1) on probability measures. When $\Omega$ is not separable the same is true if you restrict to $\tau$-additive measures. See Bogachev's Measure Theory vol 2, Theorem 8.3.2. I believe Dudley's book also treats some of this.


The weak* dual of an infinite-dimensional Banach space is never metrizable (that is, there does not exist a metric -- not necessarily translation invariant -- inducing the weak* topology). Indeed, otherwise the $0$-neighborhood filter of the weak* topology would have a countable base and it would thus be a metrizable locally convex space. The theorems of Banach-Steinhaus (or the barrelledness of the Banach space) and Alaoglu then imply that bounded sets of this metrizable locally convex topology are complete and hence, the weak* dual would be a Frechet space. The closed graph theorem implies that the weak* topology is the same as the one induced by the dual norm which is not true because every weak* neighborhood of $0$ contains a co-finite-dimensional subspace.

  • $\begingroup$ It seems easier to argue that $X^\ast$ has uncountable dimension as a vector space. Every weak*-neighborhood contains a linear subspace of finite codimension, so the intersection of countably many $0$-neighborhoods contains a subspace of countable codimension, in particular it can't be reduced to $\{0\}$. $\endgroup$
    – Martin
    Jan 30, 2013 at 12:07

If you need a reference, there is a paper that covers your question.

Varadarajan (1958): If $\Omega$ is a separable metric space, then the topology of weak convergence on $\mathcal{M}(\Omega)$ is metrizable if and only if the weak convergence and norm topologies on $\mathcal{M}(\Omega)$ coincide.

This condition is obviously violated if $\Omega$ is the unit interval (or any other uncountable separable metric space).

The same paper also shows that the weak convergence topology on the subset $\mathcal{M}^+(\Omega)$ of positive measures is metrizable.


As mentioned in several replies, you require that the compact space be metrisable for a positive answer to your question. Then the claim is, in fact, true for bounded sequences. The natural setting for this question is the Banach-Dieudonné theorem. If $E$ is a separable Banach space, then, as has been mentioned in previous answers, the unit ball of the dual space is metrisable under the weak star topology. The natural topology on the dual is then the finest locally convex topology on the whole space which agrees with the latter on this ball $—$it is even the finest such topology (i.e., not necessarily locally convex or even linear) and can also be characterised as the topology of uniform convergence on compact subsets of $E$. This topology has several nice properties $—$ it is complete and has the same bounded subsets as the norm topology $—$but it is not metrisable. However, its restrictions to the bounded sets of $E'$ are metrisable. This is the essential content of the above-mentioned result. The consequence which is relevant to the question posed is the fact that it has the same bounded convergent sequences as the weak star topology on $E'$ defined by a dense countable subset of $E$ and this it is metrisable. In order to answer the question posed, it suffices to specialise to the case where $E$ is $C(K)$ with $K$ compact and metrisable.

I should perhaps mention that the theorem of Banach-Dieudonné holds for any (i.e., not necessarily separable) Banach (or even, in suitable form, Fréchet) space and this provides information for the case where $K$ is not metrisable.

  • $\begingroup$ There seems to be a TeX error in the above answer but it is not detectable on my screen which makes it ipossible for me to correct it. $\endgroup$
    – jbc
    Jan 30, 2013 at 16:30

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