# Image measure space

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

• @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 – Robby S. Jan 13 at 20:38

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 $$\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$$.