# Duality pairing between finite signed measures and bounded continuous functions

Let $$E$$ be a metric space, $$C_b(E)$$ denote the set of real-valued bounded continuous functions on $$E$$ and $$\mathcal M(E)$$ denote the set of finite signed measures on $$\mathcal B(E)$$.

How can we show that $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E)$$ is a duality pairing between $$C_b(E)$$ and $$\mathcal M(E)$$? Do we need impose further restrictions on $$E$$ (e.g. completeness and/or separability) and/or $$\mathcal M(E)$$ (e.g. regularity of the measures)?

We need to show that \begin{align}\forall f\in C_b(E)\setminus\{0\}&:\exists\mu\in\mathcal M(E)&:\langle f,\mu\rangle\ne0\tag1;\\\forall\mu\in\mathcal M(E)\setminus\{0\}&:\exists f\in C_b(E)&:\langle f,\mu\rangle\ne0\tag2.\end{align}

$$(1)$$ Should be easy. If $$f\in C_b(E)\setminus\{0\}$$, there is a $$x\in E$$ with $$f(x)\ne 0$$. Now we can choose $$\mu$$ to be the Dirac measure $$\delta_x$$ concentrated at $$x$$ and obtain $$\langle f,\mu\rangle=f(x)\ne0$$.

But how can we show $$(2)$$?

## 2 Answers

To put this result into context, I will show how to deduce it from the big theorems. (This is not nearly as elementary as Kavi's proof.)

Unfortunately, different authors use different notation / terminology, so let me set up a few definitions first.

Definition. A finitely additive signed measure (or charge) is a function $$\mu : \mathcal A \to [-\infty,\infty]$$ with the following properties:

• $$\mu(\varnothing) = 0$$;
• $$\mu$$ assumes at most one of the values $$-\infty$$ and $$+\infty$$;
• $$\mu$$ is finitely additive.

Comparted to signed measures, the requirement of $$\sigma$$-additivity is weakened to finite additivity.

If $$\Omega$$ is a topological space, let $$\mathcal A_\Omega$$ and $$\mathcal B_\Omega$$ denote respectively the algebra and the $$\sigma$$-algebra generated by the open (or closed) sets of $$\Omega$$. Then $$\mathcal A_\Omega \subseteq \mathcal B_\Omega$$, and $$\mathcal B_\Omega$$ is the $$\sigma$$-algebra generated by $$\mathcal A_\Omega$$, so every measure on $$B_\Omega$$ is uniquely determined by its values on $$A_\Omega$$.

Definition. Let $$\Omega$$ be a topological space. We say that a finitely additive signed measure $$\mu$$ on $$\mathcal A_\Omega$$ is regular if for every $$A \in \mathcal A_\Omega$$ 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. Different authors use different notions of regularity. In particular, sometimes the closed sets $$F \subseteq A$$ are replaced by compact sets.

We will use the following well-known results:

Theorem. Let $$\Omega$$ be a normal Hausdorff space. Then $$C_b(\Omega)'$$ is isometrically isomorphic to the space $$rba(\Omega)$$ of all regular, finitely additive signed measures of bounded variation on $$\mathcal A_\Omega$$, equipped with the total variation norm.

See [DS58, Theorem IV.6.2 (p.262)] or [AB06, Theorem 14.10 (p.495)].
For general topological spaces, see this question on MathOverflow.

Lemma. Every finite signed Borel measure on a metric space is regular.

See [DS58, Exercise III.9.22 (p.170)] or [AB06, Theorem 12.5 (p.436)], among others.
(Side note: this is not true for the other notion of regularity, with compact sets instead of closed sets, as can be seen from this answer on MathOverflow.)

It follows that the space $$\mathcal M(\Omega)$$ of finite signed Borel measures is a subspace of $$rba(\Omega) \cong C_b(\Omega)'$$. To complete the proof, note that every normed space $$X$$ separates points on every subspace of $$X'$$: if $$\varphi(x) = 0$$ for all $$x \in X$$, then $$\varphi = 0$$.
(More generally, for a bilinear pairing $$\langle E , F \rangle$$ to be non-degenerate, so that it is a proper dual pairing, it is necessary and sufficient that the induced maps $$E \to F^*$$ and $$F \to E^*$$ are injective.)

References.

[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.

• The reason why $\mathcal M(E)$ (from the question) is only a subspace of your space $rba(E)$, is that I've defined $\mathcal M(E)$ to be a set of finite signed measures, which hence necessarily have bounded variation, but your $rba(E)$ might contain signed measures of bounded variation which take either the value $\infty$ or $-infty$, right? Or is there more on that? Commented Dec 7, 2020 at 10:52
• @0xbadf00d no, the point is that $rba(\Omega)$ contains a larger class of set functions, namely the finitely additive measures. Compared to $\mathcal M(\Omega)$, the $\sigma$-additivity of measures is weakened to finite additivity. Clearly bounded variation implies finite (bounded, in fact), so everything in $rba(\Omega)$ is finite. However, I don't think the converse (finite implies bounded variation) is true for finitely additive measures; I think you need $\sigma$-additivity for that. Commented Dec 7, 2020 at 22:41

If $$C$$ is any closed set and $$U_n=\{x: d(x,C) <\frac 1 n\}$$ then there exist continuous functions $$f_n : E \to [0,1]$$ such that $$f_n(x)=1$$ for all $$x \in C$$ and $$f_n(x)=0$$ if $$x \notin U_n$$. Since $$\int f_n d\mu=0$$ for all $$n$$ we get $$\mu (C)=0$$. If we write $$\mu$$ as $$\mu_1-\mu_2$$ where $$\mu_1$$ and $$\mu_2$$ are positive finite measures then we get $$\mu_1(C)=\mu_2(C)$$ for any closed set $$C$$. You can use $$\pi -\lambda$$ Theorem to show that $$\mu_1=\mu_2$$.

• What exactly have you shown? Did you start with the contrary assumption that there is a $\mu\in\mathcal M(E)\setminus\{0\}$ such that $\langle f,\mu\rangle=0$ for all $f\in C_b(E)$? I've read that the "regularity of the measures yields that $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing", but do we really need regularity at some point? Commented Nov 26, 2020 at 9:19
• There are different types of regularity. Any finite Borel measure on metric space has the regularity property $\mu (A)=\sup \{\mu (C): C \text {closed}, C \subset A\}$. @0xbadf00d Commented Nov 26, 2020 at 9:25
• How do you define the notion of being a "Borel" measurable? Some authors imply mean any measure on a Borel $\sigma$-algebra by that term; others mean that $\mu(K)<\infty$ for every compact $K$. In any case, did you use $\mu(A)=\sup\{\mu(C):C\text{ closed },C\subseteq A\}$ at some point? Commented Nov 26, 2020 at 9:29
• Your question is about finite signed measures. Any such measure defined on the Borel sigma algebra has the regularity property I have stated. (BTW Borel measure and Borel measurable are different terms. You cannot talk about a measure being Borel measurable). @0xbadf00d Commented Nov 26, 2020 at 9:37
• @0xbadf00d thanks for bringing this to my attention. I agree that this answer is a little short on the details, but it seems to be correct. To get existence of the functions $f_n$, use Urysohn's lemma. To get $\mu(C) = 0$, use the dominated convergence theorem (also works for signed measures; e.g. this answer), since $f_n$ converges pointwise to $\chi_C$. The case $C = \varnothing$ or $C = E$ deserves separate treatment. For the final step, see typical application of $\pi$-$\lambda$ theorem). Commented Dec 7, 2020 at 7:35