For which $c\in [3,\infty)$ there does exist a measure space $(X,\mathcal{S},\mu)$ such that $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup[3,c]$ I am trying to find out for which $c\in [3,\infty)$ there exists a measure space $(X,\mathcal{S},\mu)$ such that $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup[3,c]$.
My solution:
(EDIT my solution is wrong, as Calvin Khor pointed out in a comment below because one could take a set $E$ which has an overlapping of measure $1$ with $[0,1]$ and $[3,c]$ such as $[0,4]$ or an interval like $[3,4.5]$ which would then sum to a number greater than $1$ but smaller than $3$. In general, if such a measure existed for some $c>4$ then we would have that there exist $E_1, E_2\in\mathcal{S}$ such that $\mu(E_1)=3,\ 4<\mu(E_2)<4.5$ so $1<\mu(E_2)-\mu(E_1)<4.5-3=1. 5$ thus $1<\mu(E_2\setminus E_1)<4.5-3=1.5$, contradiction. So, the claim is true for $c=4$ (as shown in other answers) and cannot be true for $c>4$).
Let $(X,\mathcal{S},\mu)=(\mathbb{R},\mathcal{B},\mu)$, where $\mu:\mathcal{B}\to [0,\infty]$ is the function $\mu(E):=|E\cap [0,1]|+|E\cap [3,c]|$, where $c\in\mathbb{R}, c\geq 3$ and $|\cdot|$ is the outer measure. Then, $\mu\Big(\emptyset\Big)=|\emptyset\cap [0,1]|+|\emptyset\cap [3,c]|=|\emptyset|+|\emptyset|=0+0=0$ and if $E_1,E_2,\dots$ are disjoint sets in $\mathcal{B}$ then $$\mu(\bigcup_{k=1}^{\infty}E_k)=|\bigcup_{k=1}^{\infty}E_k\cap [0,1]|+|\bigcup_{k=1}^{\infty}E_k\cap [3,c]|=\sum_{k=1}^{\infty}|E_k\cap [0,1]|+\sum_{k=1}^{\infty}|E_k\cap [3,c]|=\sum_{k=1}^{\infty} \Big( |E_k\cap [0,1]| + |E_k\cap [3,c]| \Big)=\sum_{k=1}^{\infty}\mu(E_k)$$ so $\mu$ is a measure and by the way it is built it gives us, as all the numbers in $[0,1]$ and $[3,c]$ for every $c\in\mathbb{R}$.
Is this correct? Are there others (perhaps simpler) solutions? Thanks.
 A: Converting comment to answer on request of OP. I will also quickly give the full solution as in the links.

*

*Your $\mu$
doesn't work, because e.g. take $c=10$
then $\mu([0,4])=2$.


*For $c=4$, you can consider  $\mu = dx+3\delta_0$ where $dx$ is Lebesgue measure on $[0,1]$.


*Show that $c\ge 4$ as follows. As $\operatorname{range}\mu = [0,1]\cup [3,c]$, we see that $\mu(X)=c$, and there exist $A,B$ with $\mu(A)=3$ and $\mu(B)=1$. If $\mu(A\cap B)=0$ then $c=\mu(X) \ge \mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B) = 4$ as claimed. Otherwise, $\mu(A\cap B) \le \mu(B)=1$ and then $$3=\mu(A)> \mu(A\setminus (A\cap B))=\mu(A)-\mu(A\cap B) \ge 3-1 = 2,$$showing that $\mu(A\setminus (A\cap B)) \notin [0,1]\cup [3,c]$,  contradicting the assumption. (source: first link)


*Show that $c \le 4$ as follows. Suppose that $c>4$. If $c<5$ then choosing $E$ with $\mu(E)=3$, note that $\mu(X\setminus E)=c-3\in(1,2)$ which is contrary to assumption. If $c\ge 5$, then instead let $\mu(E)=c-2\ge 3$. Then $\mu(X\setminus E)=c-(c-2)=2$, which is again contrary to assumption. (variant of argument in second link )


*The above assumes $c\in\mathbb R$, but $c=\infty$ is also possible, e.g. $dx|_{[0,1]} + \mu$ where $dx|_{[0,1]}$ is Lebesgue measure on $[0,1]$ and $\mu(\{x\})=x$ for $x\ge 3$ and zero otherwise. (See second link.)
