# Weak*-convergence of signed measures

Let $$X$$ be a compact Hausdorff space and let $$M(X)$$ denote the space of signed measures that is naturally dual to $$C(X)$$, the space of continuous functions on $$X$$. I am interested whether the following condition:

$$\nu_n(O) \to \nu(O)$$

for every open set $$O\subset K$$ is sufficient for weak* convergence of $$\nu_n$$ to $$\nu$$, at least when $$X$$ is zero-dimensional but not necessarily metrizable. Thank you.

• By the outter regularity, this condition implies that $\nu_n$ is strongly convergent to $\nu$, doesn't it? – André Porto Oct 21 '18 at 15:33
• @AndréPorto, mustn't we have $\sup_O |\nu_n(O) - \nu(O)|\to 0$ for strong convergence? – user3522356 Oct 21 '18 at 15:34
• So, my answer lacks some arguments, so I deleted it. It is true for positive measures, though. But I don't know for signed measures – André Porto Oct 21 '18 at 16:35
• This is essentially the same as asking if this condition implies that $\nu_n$ is bounded, that is, $\sup |\nu_n|(X)<\infty$, since the claim is almost trivial in that case ("see" my failed attempt at an answer), and obviously this boundedness is necessary for weak $*$ convergence. – Christian Remling Oct 21 '18 at 18:14
• To illustrate why this is tricky, the following non-example is interesting. Let $X = \mathbb{N} \cup \{\infty\}$, and let $\nu_n = n(\delta_{n+1} - \delta_n)$ with $\nu = 0$. This sequence doesn't converge weakly. It doesn't satisfy the hypothesis either - take $O$ to be the set of odd integers - but it is not so clear how to "produce" that set. – Nate Eldredge Oct 21 '18 at 18:33

The answer is positive if the sequence $$(\nu_n)$$ is bounded; please see Theorem IV.9.15 in Dunford/Schwartz, vol. 1. PS: I just notice that this was already observed by Christian a couple of minutes ago.