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!
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 $A\ne B$).