Duality pairing between finite signed measures and bounded continuous functions

Question

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 $(2)$?

Answer 1

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.

Answer 2

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 $\mu_1=\mu_2$.