To keep things simple, we assume $X$ to be a polish space (think of $X$ as $\mathbb{R}^n$ for example). Let's denote with $P(X)$ the space of all Borel probability measure on $X$. We say $\{\mu_n\}\subset P(X)$ converges weakly to $\mu\in P(X)$, denoted by $\mu_n\Rightarrow \mu$ if

$$\int fd\mu_n\to\int fd\mu,\forall f\in C_b(X)$$

where $C_b(X)$ is the space of all bounded continuous real valued functions. Let this definition be seen as the purely probabilistic one.

From functional analysis we also have a concept of weak convergence and weak topology. Let $E$ be a Banach space and denote by $E'$ the dual. Then by considering the family $\{\phi_f:f\in E'\}$, where $\phi_f:E\to\mathbb{R}$ is the linear functional $\phi_f(x):=\langle f,x\rangle$, the weak topology on $E$ is the coarsest topology which makes all $\phi_f$ continuous. One can prove $x_n\to x$ weakly (in weak topology) if and only if $\phi_f(x_n)\to\phi(x)$ for all $f\in E'$.

Since $X$ is Polish, we have that $P(X)$ is Polish too. Now I can define a continuous linear functional $\phi_f: P(X)\to\mathbb{R}$ for $f\in C_b(X)$ by

$$\phi_f(\mu):=\int fd\mu$$

Infact $\phi_f\in P(X)'$. Therefore we have $C_b(X)\subset P(X)'$.

My questions are

$\textbf{1. Question}$ Is the dual $P(X)'$ known? Is it isomorphic to a well known space? Is $C_b(X)$ a proper subspace?

$\textbf{2. Question}$ The notation of weak convergence in the probabilistic sense ($\mu_n\Rightarrow \mu$) is weaker than the weak convergence in the functional analytical sense. What I mean is: If $\mu_n\to\mu$ in the weak toplogy, i.e. $\langle f,\mu_n\rangle \to \langle f,\mu\rangle $ for all $f\in P(X)'$ this implies $\mu_n\Rightarrow \mu$ since $C_b(X)\subset P(X)'$. Is this the reason why one calls $\mu_n\Rightarrow \mu$ weak convergence, or is there any other reason?

  • $\begingroup$ In most cases, $C_b(X)$ is dense in $P(X)'$, and so $\mu_n \to \mu$ weakly iff $\int fd\mu_n \to \int fd\mu$ for all $f\in C_b(X)$. $\endgroup$ Commented Oct 4, 2013 at 13:15
  • $\begingroup$ @PrahladVaidyanathan Interesting! What does most cases mean? Can you provide a reference? $\endgroup$
    – math
    Commented Oct 4, 2013 at 13:18

4 Answers 4


"Weak convergence of measures" is a misnomer. What it really means is that the space of measures is identified, via Riesz representation, with the dual of some space of continuous functions, and this gives us weak* topology on the space of measures. People just don't like saying that their measures converge weak-star-ly, or putting a lot of asterisks in their texts.

Folland writes in Real Analysis, page 223:

The weak* topology on $M(X)=C_0(X)^*$ ... is of considerable importance in applications; we shall call it the vague topology on $M(X)$. (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 ours, since $C_0(X)$ is rarely reflexive.

  • $\begingroup$ @DavideGiraudo The thing I don't get is, $M(X)$ is the space of complex random measure. So since $X$ is assumed to be polish, we have $P(X)\subset M(X)$. Therefore also $P(X)\subset C_0(X)^*$. It would be clear, if $P(X)=C_b(X)^*$. Is the dual of $C_b(X)$ known, for particular nice $X$? $\endgroup$
    – math
    Commented Oct 5, 2013 at 8:47
  • 1
    $\begingroup$ When paired with $C_b(X)$, as probability folks tend to do, it's probably more accurate to say it's the weak-* convergence in $C(\beta X)$, where $\beta X$ is the Stone-Cech compactification of $X$. $\endgroup$
    – Michael
    Commented Oct 6, 2013 at 10:09

Observe that the notation $P(X)'$ does not make sense as $P(X)$ is not a linear space. But you can topologize $P(X)$ in many ways:

  1. as a subspace (in the topological sense) of $C_0(X)'$ with the norm topology, the distance between any two probabilities $\mu$ and $\nu$ being $\sup\big\{\big\vert\int f\,d\mu - \int f\,d\nu\big\vert:\;f\in C_0(X),\;\Vert f\Vert\leq1\big\}$
  2. as a subspace of $C_0(X)'$ with its weak* topology, under which $\mu_n\rightarrow \mu$ if and only if $\int f\,d\mu_n\rightarrow\int f\,d\mu$ for all $f\in C_0(X)$.
  3. as a subspace of $C_b(X)'$ with the norm topology, the distance between any two probabilities $\mu$ and $\nu$ being $\sup\big\{\big\vert\int f\,d\mu - \int f\,d\nu\big\vert:\;f\in C_b(X),\;\Vert f\Vert\leq1\big\}$
  4. as a subspace of $C_b(X)'$ with its weak* topology, under which $\mu_n\rightarrow \mu$ if and only if $\int f\,d\mu_n\rightarrow\int f\,d\mu$ for all $f\in C_b(X)$. This is what probabilists usually call weak convergence.

If K denotes the Stone-Cech compactification βX of X, a description of M(K)' -viewed as the second dual of C(K) can be obtained by using the Arens product, which in the case of commutative C-* algebras implies that M(K)' is isometrically isomorphic to $C(\tilde K)$ -where $\tilde K$ is a compact (Stonean) space, called the hyperstonean envelope of $K$. There is a reecnt book devoted to such issues: H.G. Dales, F.K.Dashiell,Jr., A.T.M. Lau, D. Strauss : "Banach Spaces of Continuous Functions as Dual Space


The following is merely a sketch and I am not too sure about it (and thus am very grateful for corrections, comments or any other inputs, but concerning your question about the dual of $P(X)$:

Your space $P(X)$ is a subspace of the space of all (signed) finite regular Borel measures, $M(X)$ which is known to be a Banach space. Note that any finite Borel measure on $X$ is automatically regular, since $X$ is Polish.

Now $M(X)$ can be seen as a large $\ell_1$-sum of $L^1$ spaces as follows: Take a maximal collection $(\mu_i)_{i\in \mathcal{I}}$ of mutually singular probability measures on $X$ (use Zorn).

Let $\nu \in M(X)$. Using the Radon-Nikodým theorem, for every $i\in \mathcal{I}$ write

$$d\nu = f_i d\mu_i + \rho$$

where $\rho$ is singular with respect to $\mu_i$. Now put

$$\nu_0 = \sum_{i\in \mathcal{I}} f_i d\mu_i$$

It follows that $\nu - \nu_0$ is singular to every $\mu_i$ and hence vanishes. Therefore,

$$ M(X) \cong \left(\bigoplus_{ i\in\mathcal{I}} L^1(\mu_i) \right)_1 $$

via the map $$ \nu \mapsto (f_i)_{i\in \mathcal{I}} $$

Consequently, $M(X)'$ can be identified with the $\ell_\infty$-sum of the $L^\infty(\mu_i)$, so the $P(X)'$ will be a quotient of this space, if I am not mistaken. Of course, this does not give too much information about the true structure.


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