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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?
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2 Answers

up vote 3 down vote accepted

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.

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A variant of your idea works without $CH$. Well-order the reals and let $X_{\alpha}$ be the first $\alpha$ points in the well-ordering. Then there is a least $\lambda$ such that $X_{\lambda}$ doesn't have measure 0. Clearly $\lambda$ is a limit ordinal and $X_{\lambda} = \bigcup_{\alpha < \lambda}X_{\alpha}$ is a counterexample. –  Simon Thomas Sep 28 '10 at 12:15
    
Simon, you and I hit on that idea simultaneously! I had added it just before you submitted your comment. –  Joel David Hamkins Sep 28 '10 at 12:30
    
Thank you! That really helped me. –  Sebastian Scholtes Sep 30 '10 at 7:25
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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.

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Thanks a lot! If anyone else is interested, this can be found in Theorem 7.12. –  Sebastian Scholtes Sep 30 '10 at 7:27
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