Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a metric space, $\Sigma_{1}$ the borel sigma algebra and $\Sigma_{2}$ the sigma algebra generated by balls (open and closed).

If $\mu$ is a probability measure on $\Sigma_{2}$ can it be extended to a measure on $\Sigma_{1}?$

share|improve this question

1 Answer 1

up vote 7 down vote accepted

Take a set $X$ of power $\aleph_1$, with the discrete metric where two distinct points have distance $1$. The balls are singletons and the whole space. The ball sigma-algebra is the countable and co-countable sets. Let countable sets have measure zero, co-countable sets have measure 1.

Now all subsets are open, so the Borel sigma-algebra is the power set. There is no extension of this measure!

share|improve this answer
Maybe I'm missing something easy, but why is there no extension of this measure to the power set? –  Joel Moreira Feb 28 '13 at 20:29
A theorem of Ulam. –  Gerald Edgar Feb 28 '13 at 20:32
Ulam, Stanislaw (1930), "Zur Masstheorie in der allgemeinen Mengenlehre", Fundamenta Mathematicae 16: 140–150. –  Gerald Edgar Feb 28 '13 at 20:37
Very nice answer. But let me add: The nonextension of the measure here amounts to the fact that $\aleph_1$ is not a real-valued measurable cardinal. This is a theorem of ZFC, but we should not expect to prove this without the axiom of choice. Indeed, it is consistent (relative to large cardinals) that one has ZF+DC plus $\aleph_1$ is a measurable cardinal (with a two-valued measure). In such a set-theoretic context, the measure actually would extend. In particular, this situation is a consequence of the axiom of determinacy, which implies that the club filter on $\omega_1$ is a measure. –  Joel David Hamkins Feb 28 '13 at 23:07

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.