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 $\Sigma_{1}?$
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1 Answer
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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!
answered Feb 28, 2013 at 19:52
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$\begingroup$ Maybe I'm missing something easy, but why is there no extension of this measure to the power set? $\endgroup$ Commented Feb 28, 2013 at 20:29
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1$\begingroup$ Ulam, Stanislaw (1930), "Zur Masstheorie in der allgemeinen Mengenlehre", Fundamenta Mathematicae 16: 140–150. $\endgroup$ Commented Feb 28, 2013 at 20:37
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1$\begingroup$ 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. $\endgroup$ Commented Feb 28, 2013 at 23:07