MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
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.

share|cite|improve this answer
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

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.