Problem set up:

Consider $C_b$, the Banach space of continuous bounded functions on $[0, \infty)$ equipped with the sup norm. Denote by $M$ the set of probability measures on $[0, \infty)$, and for $r > 0$ denote by $M_r$ the set of probability measures supported on $[r, \infty)$. We will consider $M$ as subsets of the continuous dual $C_b^*$ in the usual way.

Consider the set $\mathcal S$ of linear functionals $L \in C_b^*$ such that there exist a sequence $r_n$ of real numbers with $r_n \to \infty$, and a sequence of probability measures $\mu_n$ with $\mu_n \in M_{r_n}$ for all $n$ such that $\mu_n \to L$ in the weak* topology.

By the Banach-Alaoglu theorem, $\mathcal S$ is nonempty.

Question: Is it possible to produce an explicit example of an element of $\mathcal S$, and the corresponding probability measures?

  • $\begingroup$ Ah yes, sorry. Edited. $\endgroup$
    – Nate River
    Jun 14, 2021 at 11:56
  • 1
    $\begingroup$ The question doesn't quite make sense: every probability measure on $[0,\infty)$ induces a member of $C_b^*$, but the converse is very far from being true. So $\mathcal S$ need not actually contain any functional induced by a probability measure. $\endgroup$ Jun 14, 2021 at 12:01
  • 3
    $\begingroup$ @NateRiver that's not the usual use of $C^0$, which is usually the bounded functions that vanish at infinity. Now those are usually denoted $C_0(X)$, which is indeed separable when $X$ is separable. However, the $C_b$ in your question is non-separable, as it contains $\ell^\infty$. $\endgroup$
    – user147609
    Jun 14, 2021 at 14:31
  • 1
    $\begingroup$ If I remember correctly, the set of probability measures on a separable metric space is sequentially closed in the weak* topology of $C_b^{\ast}$, so $\mathcal{S}$ would be empty, if you didn't allow nets. I don't have a reference, though, so take it with a grain a salt. $\endgroup$ Jun 14, 2021 at 14:34
  • 1
    $\begingroup$ @MateuszWasilewski That is correct, the space of measures on the real line is w* sequentially complete; see e.g. Bogachev, Measure Theory (2007), section 8.7. So the set $\cal S$ as defined in the question is empty. $\endgroup$
    – user95282
    Jun 16, 2021 at 10:50

1 Answer 1


(reading "sequence" as "net", as suggested in the comments)

Well, $C_b(\mathbb{R}^+) \cong C(\beta\mathbb{R}^+)$, so any such $L$ will arise from a probability measure on the Stone-Cech remainder $\beta\mathbb{R}^+ \setminus \mathbb{R}^+$. You need some choice principle to know this set is nonempty, so no example can be very explicit. (I think it's consistent with ZF that there are no free ultrafilters on $\mathbb{R}^+$.)

If you're willing to take ultrafilters as "explicit", then evaluation on a free ultrafilter on $\mathbb{Z}^+$ would be an example.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .