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Is it possible to come up with a measure space $(X, \mathcal{M}, \mu)$ such that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$ , where $\Bbb Q^+$ denotes positive rationals

I suspect yes, although I have no idea how to construct such a measure space like this

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  • $\begingroup$ @DonThousand if X is the empty set, then I suppose M is also empty, so {μ(E)∣E∈M} is empty and how could that be [0,+∞]\Q+?, I'm surely doing a bad reasoning $\endgroup$ – Robby S. Jan 13 at 20:38
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There is no measure space $(X, \mathcal{M}, \mu)$ such that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$ (where $\Bbb Q^+$ denotes the positive rational numbers).

Proof:

Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$.

By a result from Sierpinski, we know that the image of an non-atomic measure is an interval (see for instance, this article in Wikipedia). In fact, from Sierpinski's result, we can easy prove that: if the image of a measure $\mu$ does not contains any interval, then for every $E \in \mathcal{M}$ such that $\mu(E)>0$, $E$ contains (at least) one atom.

Since $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$, we can take a $E \in \mathcal{M}$ such that $0<\mu(E)<+\infty$. Let $A \subseteq E$ be an atom. We have that $0<\mu(A) <+\infty$.

Now take $r\in \Bbb Q$ such that $\mu(A)< r < 2 \mu(A)$. So, we have that $$0< r-\mu(A) < \mu(A)$$ Since $\mu(A) \notin \Bbb Q$ and $r\in \Bbb Q$, we have that $ r-\mu(A) \in (0, +\infty) \setminus \Bbb Q^+$. So there is $B \in \mathcal{M}$, such that $\mu(B) = r-\mu(A)$.

Since $A$ is an atom and $\mu(B) < \mu(A)$, we have that $\mu(A\cap B)=0$. So $$\mu(A \cup B) = \mu(A) + \mu(B)= \mu(A) + r-\mu(A) = r \in \Bbb Q$$ Contradiction to the assumption that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$.

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