# Is the range of a (nonnegative or signed) measure a closed set?

Halmos showed that the range of a non-negative, finite measure is a closed subset of real numbers.

1. Is this true for non-negative, even infinite measures?

2. Is this true for signed measures? If so, can it be proved without the Hahn decomposition theorem?

My reason for asking is that if we knew the range of a signed measure to be closed, we could deduce the existence of sets of maximal and minimal measure, yielding a very quick proof of the Hahn decomposition theorem.

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For 2 the answer is yes from the Hahn decomposition. But I think it is a good question whether we can do this without the Hahn decomposition. – Gerald Edgar Mar 14 '14 at 13:50

Infinite, non-negative measures: no.

Indeed, on the set $\{1,2,3,\cdots\}$ perhaps you can find values for $\mu(\{n\})$ to get a non-closed set for the range of the measure.

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Quote from "Vector measure", chapter IX (Diestel, Uhl): "One of the most beautiful and best-loved theorems of the theory of vector measures is the Liapounoff Convexity Theorem which states that the range of a non-atomic vector measure with values in a finite dimensional space is compact and convex".

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Gerald Edgar has answered the first part.

For the second, Gerald Edgar's example shows that the range of an extended signed measure need not be closed, so we consider the question for finite signed measures. It is enough to show the result for non-atomic measures: If $\mu$ is a signed measure with atoms of arbitrarily small measure, we are done. Otherwise, $\mu$ has a finite number of atoms, and we may apply the following argument to the restriction of $\mu$ to the set where it is non-atomic.

So suppose we have a non-atomic signed measure $\mu$ and let $Z$ be a measurable set of non-zero measure. Since $Z$ is not an atom, we may choose a subset $A$ of $Z$ of non-zero measure, and with $\mu(A)$ different from $\mu(Z)$. Let $E_0$ be whichever of $A$ and $Z-A$ maximizes $|\mu(\cdot)|$. Let B be a subset of $Z-E_0$ of non-zero measure with $\mu(B)$ different from $\mu(Z)$ and let $E_1$ be whichever of B and $Z-E_0-B$ maximizes $|\mu(\cdot)|$. Define the sequence $E_0, E_1, E_2,\cdots$ inductively in this way. Let $Z_n = \cup_{i=0,\cdots,n}E_i$. Since $\mu$ is finite, $\mu(Z_n)$ converges as $n\to\infty$. In particular, $|\mu(E_n)|\to 0$. $$2|\mu(E_n)|\ge |\mu(E_n)| +|\mu(Z-Z_{n-1}-E_n)| \ge |\mu(Z-Z_{n-1})|,$$ so $\mu(Z_n)\to \mu(Z)$, and hence the range of $\mu$ is closed.

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