Let $X$ be a non-empty set, let $\mathcal{A}$ be an algebra of sets on $X$, and let $\mu$ be a finite content on $(X,\mathcal{A})$. Can $\mu$ be extended to a content on $2^X$ (i.e. on $X$'s power-set)?
* As far as I know, the Lebesgue measure on $\mathbb{R}$ (and hence the Jordan content on $\mathbb{R}$) can be extended to a content on $2^{\mathbb{R}}$, but can this be done in general?
Let $\ell^\infty(X)$ be the Banach space of bounded functions on $X$ equipped with the $\sup$ norm. Let $Y$ be the subspace spanned by the set $\{ \chi_A : A \in \mathcal{A} \}$ where $\chi_A$ is the indicator function of $A$.
Notice that a finite content $\mu$ on $(X, \mathcal{A})$ defines a linear functional on $Y$ by integration, which I will denote by $\phi$. That is, for $f \in Y$, $\phi(f) = \int f d \mu$. In particular, this gives us that $\phi$ is bounded since the usual properties of the Lebesgue integral for simple functions are true for integration of simple functions against a finite content.
By the Hahn-Banach Theorem, we can extend $\phi$ to a bounded linear functional $\tilde{\phi}$ on all of $\ell^\infty(X)$. You can then check that $\tilde{\mu}(A) = \tilde{\phi}(\chi_A)$ defines a finite content which extends $\mu$ to the power set of $X$.