Problems understanding dual of $C_b[0,1]$ I know that the dual of $C_b[0,1]$ is the space of finitely additive regular measures. But $C_b[0,1]$ is the same space is $C[0,1]$ and suddenly the dual shrinks to only the $\sigma$-additive measures. Can someone explain this?
Moreover, I have the same problem with $C_b(\mathbb R)$. We have $C_0(\mathbb R)$ contained as a subspace, why should the dual of $C_0$ be smaller then?
 A: Concerning your first question, there is a theorem due to Alexandroff, which says that if $X$ is a compact topological space and
 $M$ is an algebra that contains all open sets, then every finitely additive regular measures $\lambda:M\to\mathbb{R}$ is countably additive. Moreover, if $X$ is a Hausdorff space, then $\lambda$ can be extended in a unique
way as a signed Radon measure to the smallest $\sigma$-algebra that contains $M$.
You can find this in Dunford-Schwartz or Fonseca-Leoni.
Concerning your second question, if you have a linear continuous functional $L:C_0(\mathbb{R})\to \mathbb{R}$ then you can represent it as $L(f)=\int_\mathbb{R}f\,d\mu$, where $\mu$ is a Radon measure. Since $C_0(\mathbb{R})$ is contained in $C_b$ or in the space of bounded functions, by the Hahn-Banach theorem you can extend $L$ to a continuous linear functional defined in those spaces. In turn the extension can be represented as an integral with respect to a finitely additive measure, say $\lambda$. All you know is that $\int_\mathbb{R}f\,d\mu=\int_\mathbb{R}f\,d\lambda$ for all $f\in C_0(\mathbb{R})$. The standard example is the   Banach limit. The idea is that you can have "nasty" functionals which, when restricted to smaller spaces, are much nicer.
