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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.

user3522356
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  • 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
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    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
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    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
  • This question has connections to an old question of mine. In particular, if the condition here had "Borel sets" instead of "open sets" then the claim would be true by the result of Darst cited in my answer (there might be an earlier result of Nikodym that suffices in this case). And if the answer to this question is yes, then it will give another counterexample to my Q2. – Nate Eldredge Oct 22 '18 at 04:09
  • In fact, it seems to me that it might be possible to adapt Darst's "gliding hump" argument to prove this statement. But it may take some thought. – Nate Eldredge Oct 22 '18 at 21:39

1 Answers1

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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.

Dirk Werner
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