Let $E$ be a metric space, $C_b(E)$ denote the space of bounded continuous functions $E\to\mathbb R$ (equipped with the supremum norm), $\mathcal M(E)$ denote the space of finite signed measures on the Borel $\sigma$-algebra $\mathcal B(E)$ (equipped with the total variation norm) and $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E).$$

I'm searching for a reference of a functional analytic proof showing that a linear functional $\varphi$ on $C_b(E)$ is continuous if and only if $$\exists\mu\in\mathcal M(E):\varphi=\langle\;\cdot\;,\mu\rangle\tag1.$$

EDIT (YC): the assertion made above is false, as has been pointed out in comments by various people, but seems to have gone unacknowledged.

All similar results (e.g. in Bogachev's Measure Theory) I've found are either treating way more general settings or establish the result in a way where I got the feeling that the arguments can be significantly simplified once one is aware of certain basic results on locally convex topologies arising from duality pairings.

In general, if $X,Y$ are $\mathbb R$-vector spaces, $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing between $X$ and $Y$ and $\sigma(X,Y)$ denotes the topology on $X$ generated by $$p_y(x):=|\langle x,y\rangle|\;\;\;\text{for }x\in X$$ for $y\in Y$, we know that for $\varphi\in X^\ast$

  1. $\varphi$ is $\sigma(X,Y)$-continuous;
  2. $\exists k\in\mathbb N:\exists y_1,\ldots,y_k\in Y:\exists c\ge0:|\varphi|\le c\displaystyle\max_{1\le i\le k}p_{y_i}$;
  3. $\exists y\in Y:\varphi=\langle\;\cdot\;,y\rangle$

are equivalent.

Maybe we need to impose further assumptions (e.g. restrict ourselves to the subspace $\mathcal R(E)$ of Radon measures in $\mathcal M(E)$), but I think we should be able to find a proof for the desired claim using the aforementioned equivalence.

Maybe a result similar to Bogachev, but in the present simpler setting, can be established: It holds $(1)$ for a Radon measure $\mu$ if and only if $\varphi$ satisfies $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact }:\forall f\in C_b(E):\left.f\right|_K=0\Rightarrow|\varphi(f)|\le\varepsilon\left\|f\right\|_\infty\tag2.$$

  • 9
    $\begingroup$ I confess to not reading all of your post, but as stated your result (1) is incorrect unless $E$ is compact. For example, take $E={\mathbb N}$ with the $\{0,1\}$-valued metric; then $C_b(E)=\ell^\infty({\mathbb N})$ and there are continuous functionals on $\ell^\infty(\mathbb N)$ which are not given by finite Borel measures on $E$ (since such measures are determined by what they do on singleton sets, and hence correspond to elements of $\ell^1(\mathbb N)$. If you replace $C_b(E)$ with $C_c(E)$ then the result is true and this is the Riesz representation theorem. $\endgroup$
    – Yemon Choi
    Dec 7, 2020 at 21:23
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    $\begingroup$ Alternatively, if $C_b(E)$ is equipped with a weaker topology (perhaps the so-called strict topology of Buck would do) then the statement (1) should become true. The intuition is that you want to rule out functionals which "only depend on what elements of $C_b(E)$ do as their supports move off to infinity" $\endgroup$
    – Yemon Choi
    Dec 7, 2020 at 21:26
  • $\begingroup$ I will add to Yemon's good points that you can also pick your axiomatic system so that $(\ell^{\infty})'=\ell^1$. You just need to give up AC, but you can keep DC. See sciencedirect.com/science/article/pii/S0019357798800396 $\endgroup$ Dec 7, 2020 at 21:48
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    $\begingroup$ @YemonChoi: Note that for your $C_c(E)$ statement to be interesting, $E$ should be locally compact. $\endgroup$ Dec 7, 2020 at 23:04
  • $\begingroup$ @NateEldredge Oops, yes I had tacitly made that assumption without even realising it $\endgroup$
    – Yemon Choi
    Dec 8, 2020 at 14:32

3 Answers 3


As I already pointed out in this answer to another recent question of yours, the dual of the Banach space $C_b(E)$ is not $\mathcal M(E)$ but the larger space $rba(E)$ of signed set functions that are regular, bounded, and additive, defined on the algebra $A_E$ generated by the open (or closed) sets of $E$. In other words, $rba(E)$ consists of all functions $\mu : \mathcal A_E \to \mathbb{R}$ satisfying the following conditions:

  • $\mu(\varnothing) = 0$;
  • $\mu$ is finitely additive (not necessarily $\sigma$-additive);
  • $\mu$ is bounded;
  • for all $A \in \mathcal A_E$ one has \begin{align*} \mu(A) &= \sup\{\mu(F) \, : \, F \subseteq A\ \text{closed}\} \\[1ex] &= \inf\,\{\mu(V) \: : \: V \supseteq A \ \text{open}\}. \end{align*} (Note: depending on the context, some authors use a different notion of regularity, where the closed sets $F$ in the supremum are replaced by compact sets. The results stated in this answer are no longer correct with this other notion of regularity.)

When equipped with the total variation norm, $rba(E)$ is isometrically isomorphic with $C_b(E)'$; see [DS58, Theorem IV.6.2 (p.262)] or [AB06, Theorem 14.10 (p.495)]. Your space $\mathcal M(E)$ embeds (isometrically) in $rba(E)$, since every finite signed Borel measure on a metric space is automatically regular; see [DS58, Exercise III.9.22 (p.170)] or [AB06, Theorem 12.5 (p.436)]. However, in general, $rba(E)$ is larger than $\mathcal M(E)$; see this answer.

The references mentioned here are the only functional analysis textbooks that I am aware of that prove this specific result, but maybe others know additional references. Note: although a proof written in the language of functional analysis might be more pleasant to read for some, I don't see how that would lead to a significant simplification. Basic results on dual pairs are not going to help you determine the dual of any particular Banach space; you still have to get your hands dirty. (The Riesz representation theorem doesn't prove itself, so to speak.)

For an overview of the norm duals of various Banach spaces related to measure theory, see [AB06, Table 14.1 (p.499)].


[DS58] Nelson Dunford, Jacob T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.

[AB06] Charalambos D. Aliprantis, Kim C. Border, Infinite Dimensional Analysis, A Hitchhiker's Guide, Third Edition, Springer, 2006.


This is a well-studied area and the two results relevant to your question (which hold for the space $X=C^b(E)$ of bounded, continuous functions on any completely regular space $E$) are:

  1. the space of bounded, finitely additive meaures on the Baire $\sigma$-algebra ($=$ the Borel algebra in the metric case) is the dual of the Banach space $X$;

  2. the space of bounded, Radon measures is the dual of $X$ with the finest locally convex topology which agrees with that of compact convergence on the unit ball.

  • 1
    $\begingroup$ Perhaps change $E$ to another letter since it was used in the question for the "space on which the functions live"? $\endgroup$
    – Yemon Choi
    Dec 8, 2020 at 14:33
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    $\begingroup$ I'm suspicious about (1). For example, if $E$ (the $E$ from the question, that is) is compact, then $C_b(E)=C(E)$ and the dual of this is the space of regular finite signed Borel measures ($\sigma$-additive, not just finitely additive). $\endgroup$ Dec 8, 2020 at 14:57
  • $\begingroup$ @Yemon Choi. Thanks—fixed $\endgroup$
    – user131781
    Dec 8, 2020 at 15:14
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    $\begingroup$ @ChristianRemling In a compact space, every regular finitely additive measure is automatically $\sigma$-additive; see [DS58, Theorem III.5.13 (p.138)] or [AB06, Theorem 12.4 (p.436)] in the references given in my answer. (However, it seems that user131781 forgot to require regularity.) $\endgroup$ Dec 9, 2020 at 7:28
  • $\begingroup$ @J.vanDobbendeBruyn: Thanks, I checked this in AB. This is surprising to me since (for example) $L^{\infty}(0,1)^*=$ finitely additive measures ($\ll$ Lebesgue measure) $\supsetneqq$ measures, and here one could take the compact set $[0,1]$ as the domain. But apparently regularity (or lack of it) makes all the difference. $\endgroup$ Dec 9, 2020 at 15:57

As noted, the dual of $C_b(X)$ is not what you thought.
A nice text on this sort of duality is:

V. S. Varadarajan, "Measures on topological spaces". (Russian) Mat. Sb. (N.S.) 55 (97) 1961 35–100.
English translation: Amer. Math. Soc. Transl. 48 (1965) pp. 161-228

He considers not only metric spaces, but the material is still interesting when restricted to metric spaces.

This is the classic discussion of funcitonals that are "$\sigma$-smooth" or "$\tau$-smooth" or "tight".

  • 2
    $\begingroup$ A nice paper but be careful: The generalized Prohorov theorem in the appendix is wrong. See. R.O. Davies, Bull. London Math. Soc. 3 (1971) 341. $\endgroup$
    – user95282
    Dec 9, 2020 at 22:06

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