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Lyapunov's theorem shows that the range of a finite-dimensional non-atomic vector measure is closed and compact. What if we do not assume the vector measure is non-atomic? Is the range still necessarily closed? (Obviously convexity may fail.)

If it e.g. a Borel measure on a topological space then the answer is yes, as in such a case we can remove a countable set of atoms $x_1,x_2,x_3,\ldots$ it's easy to show that range of the purely atomic part is closed, and then to apply Lyapunov's theorem to the non-atomic part.

But in stranger spaces...?

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  • $\begingroup$ It does not matter what the $\sigma$-algebra is. Just partition the measurable space into the disjoint union of at most countably many atoms and a nonatomic subspace. $\endgroup$
    – Iosif Pinelis
    Commented Sep 9 at 16:08
  • $\begingroup$ OK, I see, but the difference is is that the atoms needn't be single points, rather more general sets. Thnx. $\endgroup$
    – John Levy
    Commented Sep 9 at 21:52

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