Let $(X,\mathcal S,\mu)$ be a measure space.
I've been asked to reason why the set \begin{equation} \{ \mu(E) : E\in\mathcal S \} = [0,1) \end{equation} cannot exist. A mapping $\mu : \mathcal S \to [0,\infty]$ is a measure, if it satisfies the conditions
- $\mu(\emptyset) = 0$ and
- $\displaystyle\mu\left( \bigcup_{k=1}^{\infty} E_k \right) = \sum_{k=1}^{\infty} \mu(E_k)$ for disjoint sets $E_k\in\mathcal S$, where $\mathcal S$ is a $\sigma$-algebra.
I've been trying to think if the disjointness of he sets $E_k$ would be of any use here. Since $\mathcal S$ is a $\sigma$-algebra:
$\emptyset\in\mathcal S$,
$E\in\mathcal S \Rightarrow X\setminus E \in\mathcal S$ and
- if $(E_k)_{k=1}^{\infty}$ is a sequence of sets where $E_k \in\mathcal S$ for all $k\in\mathbb Z^+$, then $\displaystyle\bigcup_{k=1}^{\infty} E_k \in\mathcal S $.
In order for $\mu(E)$ to map into an interval in the first place, it would have to consist of disjoint sets $E_k$, as in $E = \bigcup_{k=1}^{\infty} E_k$. Even if these sets were not disjoint, another union of disjoint sets $A_k$ could be formed from them, i.e. if $E = \bigcup_{k=1}^{\infty} E_k, E \in \mathcal S$, take the differences \begin{align} A_1 &= E \setminus \bigcup_{k=2}^{\infty} E_k,\\ A_2 &= E \setminus A_1\setminus \left(\bigcup_{k=3}^{\infty} E_k\right),\\ A_3 &= E \setminus A_1\setminus A_2 \setminus \left(\bigcup_{k=4}^{\infty} E_k\right),\\ &\vdots\\ A_k &= E \setminus A_1\setminus A_2 \setminus \cdots \setminus A_{k-1}\setminus\left(\bigcup_{k=k+1}^{\infty} E_k\right)\\ &\vdots \,. \end{align} Therefore the non-disjointness of any union of sets $E_k$ is not a problem. A measure can always be taken even if the original set were not a collection of disjoint sets.
For the measure to be half open on the other hand, well... I would probably need to brush up on set theory but this might be the preventing factor. If all we know is that the set $E = \bigcup_{k=1}^{\infty} A_k$, where the sets $A_k$ were disjoint, all we can really say is that $\bigcap_{k=1}^{\infty} A_k = \emptyset$, but to me that says nothing about the openness or closedness of the union $E$. I know that $\emptyset$ is defined to be open, and therefore its complement is closed and any union of closed sets is closed, but I'm not sure this is getting me anywhere.
Any ideas or hints?
If $\{\mu(E)\mid E\in\mathcal S\}=[0,1)$ then we are dealing with a finite measure, and $\mu(X)$ will serve as maximal element of the collection.
However the set $[0,1)$ has no maximal element, so a contradiction is found.