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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.

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  • $\begingroup$ By the outter regularity, this condition implies that $\nu_n$ is strongly convergent to $\nu$, doesn't it? $\endgroup$
    – André Porto
    Commented Oct 21, 2018 at 15:33
  • $\begingroup$ @AndréPorto, mustn't we have $\sup_O |\nu_n(O) - \nu(O)|\to 0$ for strong convergence? $\endgroup$
    – user3522356
    Commented Oct 21, 2018 at 15:34
  • $\begingroup$ 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 $\endgroup$
    – André Porto
    Commented Oct 21, 2018 at 16:35
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    $\begingroup$ 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. $\endgroup$
    – Christian Remling
    Commented Oct 21, 2018 at 18:14
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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Oct 21, 2018 at 18:33

1 Answer 1

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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.

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