# Measures with bounded total variation norm compact in $M(X)$?

Let $X$ be a separable, metric, compact space. (e.g. an interval in $\mathbb{R}$ like $[0,10]$).

Let $M(X)$ be the set of all finite signed measures over $X$ with weak-*-topology (in probability theory also called weak-topology), e.g. dual to bounded continuous functions over $X$.

Then define $A= \left\{ \mu \in M(X) : |\mu|(X) \leq a \right\}$ for $a>0$, and $|\mu|(X)$ the total variation norm.

Now my question:

Is the set $A$ compact in $M(X)$ ?

• Yes. math.stackexchange.com/questions/479332/… provides a reference. – Thomas Rippl Mar 7 '14 at 8:12
• This reference there leads only to sequential compactness and not to compactness (because $M(X)$ is not metrizable)... Hence I think it does not answer this question here... – mimi Mar 13 '14 at 11:48

Yes. The Banach-Alaoglu theorem states that the unit ball of the dual of any Banach space is weak$^*$-compact and $M(X)$ equipped with the total variation norm is nothing but the dual of $C(X)$. By scaling the result holds for any $a>0$.