Is there a measure space $(X,\mathcal M, m)$ such that $\{m(E) \mid E \in \mathcal M\} = \Bbb Q_{\geq 0} \cup \{+\infty\}$? I have in mind the following question:

Is there a measure space $(X,\mathcal M, m)$ such that the range of $m$ satisfies $S:=\{m(E) \mid E \in \mathcal M\} = \Bbb Q_{\geq 0} \cup \{+\infty\}$?

(I would also accept a space where $\Bbb Q_{\geq 0} \cup \{+\infty\}$ is replaced by $\Bbb Q_{\geq 0}\,$.)

An idea would be to take $X=\Bbb N$ and define $m(\{n\}):=r_n$ the $n$-th positive rational number. But then $m(\{k(n) \in X \mid r_n=1/n^2, n\geq 0\})=\pi^2/6$ is not rational. So the measurable sets corresponding to $1/n^2$ shouldn't be disjoint.
To avoid this, we could demand that some fixed element $x_0$ belongs to every non-empty measurable set. But this is not possible since $\mathcal M$ is a $\sigma$-algebra, in particular it is closed under taking complements.
Similarly, if $(x_n)$ is any sequence of positive rational numbers that converges to $\sqrt 2$, the measurable sets corresponding to $x_n$ shouldn't included one in another (to avoid a chain).
I could replace $\pi^2/6$ and $\sqrt 2$ by any positive real number (since $\Bbb Q$ is dense in the reals)!
Therefore, my intuition is that such a measure space can't exist. Actually, I believe that the set $S$ defined above should be closed in $\Bbb R \cup \{+\infty\} \cong S^1$ (and even "closed under taking series" with elements in $S$). But I'm unsure if this is true, and how to prove it.
Any comment would be appreciated!
 A: The answer to your question is no, and this is not hard. But first, for the record we should point out that one of your conjectures about this is false:
The range of a measure need not be closed.
In fact, although the answer to your question is no, there is a measure with range equal to $$\{0,\infty\}\cup(\Bbb Q\cap[1,\infty)).$$ Say $(r_1,r_2,\dots)$ is an enumeration of the rationals greater than or equal to $1$, and define a measure on $\Bbb N$ by $$\mu(\{n\})=r_n.$$Then every $r_n$ is in the range of $\mu$. If $E\subset\Bbb N$ is finite and nonempty then $\mu(E)$ is a rational greater than or equal to $1$, while if $E$ is infinite then $\mu(E)=\infty$.
So that's interesting. But the range of a measure cannot be all the non-negative rationals plus $\infty$. For example:
Theorem If $\mu$ is a measure such that for every $\delta>0$ there exists $E$ with $0<\mu(E)<\delta$ then the range of $\mu$ is uncountable.
Proof: Choose sets $F_n$ with $\mu(F_n)>0$ and $$\mu(F_{n+1})<\mu(F_n)/10.$$Let  $$E_n=F_n\setminus\bigcup_{k=n+1}^{\infty}F_k\quad(n\ge2).$$Then the $E_n$ are disjoint, $\mu(E_n)>0$ and $$\mu(E_{n+1})<\mu(E_n)/3.$$For $A\subset\Bbb N$ let $$S_A=\bigcup_{n\in A}E_n.$$Then $$\mu(S_A)=\sum_{n\in A}\mu(E_n),$$and the fact that $\mu(E_{n+1})<\mu(E_n)/3$ shows that those sums are all distinct (that is, $\sum_A\ne\sum_B$ if $A\ne B$).
