While trying to answer this question, I've come across the notion of the weak*-topology on a set of probability measures. I'd like some clarification about what this means.

More specifically, let $(\Omega, \mathcal{F})$ be a measurable space. We don't assume that $\Omega$ has any metric or topological structure. What does it mean to equip the set $\mathcal{M}$ of probability measures on this space with the weak*-star topology?

I understand that the weak*-topology is the weakest topology on the dual space $V'$ of a normed vector space $V$ that makes the evaluation functionals defined by $\lambda_f(\phi) = \phi(f)$, $\phi \in V'$ and $f \in V$, continuous. What I don't understand is how $\mathcal{M}$ can be equipped with this topology as it's not a vector space.

From what I've read, I think that measures in $\mathcal{M}$ are being identified with linear functionals on a space of measurable functions. For instance, $P \in \mathcal{M}$ gives rise to a linear functional $\phi$ on the normed linear space of bounded $\mathcal{F}$-measurable functions, equipped with the $\sup$-norm, by $\phi(f) := \int f dP$. Is something like this correct? Which underlying vector space of measurable functions should be used?

I would appreciate if someone could please sketch the relevant theory for me and/or refer me to a comprehensive textbook treatment of this topic.

Addendum. My current understanding of this topic is summarized as part of my attempt to answer my own question in the link above.


2 Answers 2


You can consider the set of probabilities over $\mathcal{F}$ as a subset of the linear space $V(\mathcal{F})$ of all finitely additive (bounded) scalar measures over $\mathcal{F}$ endowed with the variation norm (see Theory of charges (K. Bhaskara Rao, M. Bhaskara Rao), chapter 7). We need to show that this space is the topological dual of another (locally convex) topological vector space.

When $\mathcal{F}$ is a Boolean algebra of subsets of $\Omega$ (which in this case does not have to be a topological space) we define $S(\mathcal{F})$ as the linear space generated by $\{\chi_A:\ A\in\mathcal{F}\}$ (the characteristic functions of the sets in $\mathcal{F}$). $S(\mathcal{F})$ is called the space of simple functions.

Now, for every $f\in S(\mathcal{F})$, $\Vert f\Vert_s:=\sup\vert f\vert<\infty$ so $(S(\mathcal{F}),\Vert\cdot\Vert_s)$ is a normed space. It is not hard to prove that the dual of $(S(\mathcal{F}),\Vert\cdot\Vert_s)$ is the space $V(\mathcal{F})$. In fact, on one hand every $\lambda\in V(\mathcal{F})$ defines the bounded linear functional $f\mapsto \int f\ d\lambda$ (for $f\in S(\mathcal{F})$), on the other, every $x^*\in S(\mathcal{F})^*$ defines on $\mathcal{F}$ the measure $\lambda_{x^*}(A):=x^*(\chi_A)$ (for $A\in\mathcal{F}$). See Topological Riesz spaces and measure theory (Fremlin) for a more complete reference.

Having showed that the set of probabilities over $\mathcal{F}$ is contained in the topological dual of the normed space $S(\mathcal{F})$, it is clear why we can talk about weak$^*$-convergence.

PS: It is worth observing that by the Stone representation Theorem for Boolean rings any Boolean ring $\mathcal{R}$ is isomorphic to the ring of clopen sets in a locally compact Hausdorff space (see Measure theory Vol III (Fremlin), or this survey by Tao). Following this line, the approach showed by Tomasz can be proved to be much closer to the one exposed so far than someone would think. It would be interesting to go through this idea.

  • $\begingroup$ This is great, thanks so much. $\endgroup$
    – aduh
    Feb 8, 2017 at 22:13

As far as I know, the right context here is that you have a locally compact Hausdorff space $X$, which give you the Banach space $C_0(X)$ of continuous functions vanishing at infinity (i.e. the set of all $x$ such that $\lvert f(x)\rvert\geq \varepsilon$ is compact, for all $\varepsilon>0$), equipped with the supremum norm.

Notice that each (signed, or even complex) regular Borel measure of bounded variation gives you a continuous linear functional on $C_0(X)$ (via integration). By Riesz-Markov theorem, the converse is also true: every continuous functional on $C_0(X)$ is given by integration with a (unique) complex, regular Borel measure of bounded variation.

Thus $C_0(X)^*$ can be identified with the space of all complex, regular Borel measures of bounded variation, and thus you can put the weak and weak-$^*$ topology on the latter. In particular, you can see the regular Borel probability measures as a subset of $C_0(X)^*$ and endow them with the appropriate subspace topology.

I haven't heard of a definition of weak-$^*$ topology on the space of measures that does not somehow reference a topology on the underlying measure space. You could find an isomorphism with a topological measure space and pull the measure through it, but I don't think the weak-$^*$ topology would be independent of the choice of isomorphism (though I think weak topology might be).

  • $\begingroup$ Thanks, this is helpful. I suppose I don't quite see why the topological structure of $X$ is absolutely essential. If you have the time and inclination, perhaps you could take a look at the post I linked to in my question. There, I elaborate more on what I said in my question here. $\endgroup$
    – aduh
    Jan 27, 2017 at 18:36
  • $\begingroup$ @aduh: The simplest argument I can think of is that for any Borel function (complex valued) function on a Polish space $X$, you can find a richer topology with the same Borel sets which makes the function continuous. Thus if weak-$^*$ topology was independent of the topology of the underlying space, it would be just the topology of pointwise convergence on Borel functions. But (at least for compact Polish spaces, if I remember correctly, I don't remember how it is in general) this is precisely the weak topology, which is strictly stronger. $\endgroup$
    – tomasz
    Jan 27, 2017 at 19:13
  • $\begingroup$ @aduh There is a hole in the argument I have put in the previous comment, because because the richer topology may a priori be no longer locally compact (I'm not sure if this can be easily fixed) and the function could not "vanish at infinity" in the richer topology. But if you want it to not depend on topology at all, then that should not be an issue. $\endgroup$
    – tomasz
    Jan 27, 2017 at 19:15
  • $\begingroup$ @tomasz I don't see why $C_0(X)*$ can be identified with space of all complex, regular borel measures. The dual space of $C_0$ doesn't necessarily contain only continuous functional. It requires the weak-topology on the $C_0$ to make those functional continuous. I think you need first put the weak topology and then conclude $C_0(X)*$ is space of all complex, regular borel measures. $\endgroup$ Aug 1, 2017 at 22:18
  • $\begingroup$ $C_0(X)$ has its own topology coming from the supremum norm. Putting a "weak topology" on a space and then defining the dual in terms of that would be circular... Besides, I'm not at all sure the dual would be what you want, anyway. $\endgroup$
    – tomasz
    Aug 2, 2017 at 1:42

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