Increasing sets Lemma for chains Let $\mathcal{V}$ be an outer measure on $X$, 
$(A_\alpha)_{\alpha\in I}$ be a chain of increasing subsets of $X$.


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*Is it true that $\mathcal{V}(\bigcup_{\alpha\in I}A_\alpha)=\sup_{\alpha\in I}\mathcal{V}(A_\alpha)$?

*If this is not true in general, are there classes of spaces $X$ and outer measures $\mathcal{V}$ (for example Hausdorff measures on metric spaces) for which this holds?

 A: One can see a counterexample easily for the reals $\mathbb{R}$ if the Continuum Hypothesis holds, for in this case the reals $\mathbb{R}$ are the union of a chain of countable sets. Simply well-order the reals in order type $\omega_1$ and for countable ordinals $\alpha$ let $X_\alpha$ be the first $\alpha$ many points in this enumeration. So every $X_\alpha$ is countable, but the union is all of $\mathbb{R}$. 
More generally, avoiding the CH assumption, let $\kappa$ be the cardinality of the smallest non-measure $0$ set $X$ of reals.  We can enumerate $X$ in order type $\kappa$, and the initial segments of this enumeration all have measure $0$, but the union is $X$, which is non-measure $0$.
The cardinal $\kappa$ used above is known as the uniformity number for Lebesgue measure in the theory of cardinal characteristics; for example, see this MO answer or Andreas Blass' handbook article.
A: This is true if $X$ is a locally compact Hausdorff space, $\mathcal{V}$ is a Radon measure on $X$ and the sets $A_\alpha$ are open. See Folland's "Real Analysis" where he proves a slightly more general result.
