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Let $(X,M,\mu)$ be a complete measure space.

Does the set $\{\mu(E)|E\in M ,\mu(E)<\infty\}$ have to be a closed subset of $R$ ?

Thank you

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See if the following argument is correct. Let $A = \{\mu(E)| E\in M, \mu(E)<\infty \}$ Since $\mu$ is continuous at $E$ if $\mu(E)$ is finite. Now, suppose that $\mu(E)<\infty$. For every $a$ in a neighborhood of $\mu(E)$, there exists a measurable set $F \in M$ st $a=\mu(F)$ is finite (such that $a=\mu(F) \in A$). Therefore, we can find a sequence $\{a_n \in A\}$, with $a_n \to \mu(E)$. Therefore $A$ is closed. Look liked that I didn't use the fact that $X$ is complete..I may be wrong.. –  user52350 Dec 9 '12 at 12:06
    
A closed subset of the set of real numbers. Ii already put the restriction that the measure of these sets is finite –  Amr Dec 9 '12 at 12:08
    
To be clear, is $\mu$ a positive measure? –  Davide Giraudo Dec 9 '12 at 12:13
    
yes This is not a signed measure –  Amr Dec 9 '12 at 12:15
    
@Randy: apply your proof to the counterexample of anonymous to see what's what. –  GEdgar Dec 9 '12 at 14:04
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2 Answers

up vote 5 down vote accepted

The answer is no. Take the positive integers (or any set containing them) and put a point mass of weight $1-\frac{1}{n}$ at every positive integer. Then $1$ is in the closure of your set but doesn't occur as the measure of any set.

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A partial answer is given by the fact that the range of a finite measure is closed. Also, the range is convex and hence, by the preceeding result, closed if the measure space is atomless.

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Thank you. This is a useful partial answer –  Amr Dec 9 '12 at 12:36
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