TITLE: Completion of spaces of measures w.r.t. weak norms
QUESTION [8 upvotes]: For the sake of concreteness denote by $M_0(X)$ the linear space of all signed Borel measures $\sigma$ with $\sigma(X)=0$ on some metric space $(X,d)$ and fix some base point $x_0\in X$. On this space define the norm
$$
\newcommand{\norm}[1]{\|#1\|}
\newcommand{\Lip}{\mathrm{Lip}}
\norm{\sigma}_0^* = \sup\{\int_X f\,d\sigma\ :\ \Lip f\leq 1,\ f(x_0)=0\}
$$
where the supremum is taken over all Lipschitz functions on $X$ and $\Lip f$ denotes the Lipschitz constant of $f$. In Bogachev's "Measure theory" (§8.10) this norm is called Kantorovich-Rubinshtein norm and it is shown that convergence in the Kantorovich-Rubinshtein metric implies weak convergence. Moreover, it is stated that $M_0(X)$ is not complete with this norm provided that $X$ is not complete. This can be seen by as follows: Assume that there exist sequences $x_k,y_k\in X$ which do not converge, but $d(x_k,y_k)\to 0$. Then define the sequence $\sigma_k = \delta_{x_k}-\delta_{y_k}$ and observe that
$$
\norm{\sigma_k}_0^* \leq d(x_k,y_k)\to 0
$$
in other words $\sigma_k\to 0$ w.r.t $\norm{\cdot}_0^*$. However, $\sigma_k$ does not convergence weakly to zero.
My question is:
What is the completion of $M_0(X)$ w.r.t. $\norm{\cdot}_0^*$?
I would like some description as a dual space or derived from some space of measures.
It feels like it should be something like a dual space of differentiable functions, or derivatives of some measure…

REPLY [11 votes]: This is known as the Arens-Eells space $AE(X)$. In the nonlinear Banach space literature it's also called the Lipschitz-free space $\mathcal{F}(X)$. It is not a dual space in general, but rather the predual of the space ${\rm Lip}_0(X)$ of Lipschitz functions which vanish at $x_0$.
In some cases it is a dual space. For instance, if $X$ is a compact metric space and $0 < \alpha < 1$, let $X^\alpha$ be the set $X$ equipped with the "Holder" or "snowflake" metric $\rho^\alpha$. In this case $AE(X^\alpha)$ is the dual of the space of "little" Lipschitz functions which are locally flat.
The space $AE(X)$ is characterized abstractly by its universal property: there is a natural isometric embedding of $X$ in $AE(X)$ (in your formulation, take the point $p$ to the Dirac measure at $p$ minus the Dirac measure at $x_0$), and for any Banach space $E$ and any Lipschitz function $f: X \to E$ which takes $x_0$ to $0$ there is a unique bounded linear extension $T: AE(X) \to E$. Moreover, $\|T\| = Lip(f)$.
You can find out all you want about this space in Chapter 3 of my book Lipschitz Algebras (second edition).