Let $\mathcal{V}$ be an outer measure on $X$, 
$(A\_\alpha)\_{\alpha\in I}$ be a chain of increasing subsets of $X$.


1. Is it true that $\mathcal{V}(\bigcup\_{\alpha\in I}A\_\alpha)=\sup\_{\alpha\in I}\mathcal{V}(A\_\alpha)$?
2. 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?



Increasing sets Lemma for chains
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](https://mathoverflow.net/questions/8972#9027) or [Andreas Blass' handbook article](http://www.math.lsa.umich.edu/~ablass/hbk.pdf).


