Properties of measures on the $\sigma$-algebra $\mathcal{S}=\{E\subset X:E\text{ is countable or }X\setminus E\text{ is countable}\}$ I am trying to find out which properties the measures $\mu$ on $(X,\mathcal{S})$, where $$\mathcal{S}=\{E\subset X:E\text{ is countable or }X\setminus E\text{ is countable}\}$$ have.
What I have done:

*

*In general we expect that $\mu(E_{countable})\leq\mu(E_{uncountable})$ since otherwise we would generate counterexamples like $X=\mathbb{R}, \mathbb{Q}\subset\mathbb{R}$ but $\mu(\mathbb{Q})>\mu(\mathbb{R})$.


*We cannot assign a non-zero constant measure to countable sets and a non-infinite measure to uncountable sets because otherwise we would generate counterexamples like the following: let $\{q_i\}_{i=1}^{\infty}$ be an enumeration of the rational numbers in $[0,1]$ and consider that $\mu(\bigcup_{i=1}^{\infty}\{q_i\})=\sum_{i=1}^{\infty}\mu(\{q_i\})=+\infty>\mu([0,1])$ but $\bigcup_{i=1}^{\infty}\{q_i\}\subset [0,1]$.
Now, all the measure I have seen on this $\sigma$-algebra up to now usually have, in accordance with what I have shown above, the form: $$\mu(E)=\begin{cases}0 & \text{if $E$ is countable} \\ \text{number }\neq 0 & \text{ if }E\text{ is uncountable}\end{cases}$$
and I also think it should be possible to prove that such measures don't have the property that $\{\mu(E):E\in\mathcal{S}\}=[0,b],\ b\in\mathbb{R}, 0<b$.
Is this correct? Are there any other properties that characterize this $\sigma$-algebra?
 A: For $\sigma$-finite measures $\mu$, there are at most countably many atoms.  That is,
$$A = \{a \in X : \mu(\{a\}) > 0\}$$
is at most countable.  Then $\mu$ has the form
$$\mu(E) = \sum_{a \in E \cap A} \mu(\{a\}) + \mu_{co}(E)$$
where $$\mu_{co}(E) = \begin{cases}0, & \text{if } E \text{ is countable} \\ c, & \text{otherwise}\end{cases}$$
and $c \geq 0$ is some constant independent of $E$.

In particular, if we fix $b \geq 0$ and set $X = \mathbb{Q},$ then $\mathcal{S} = 2^\mathbb{Q}$. By enumerating $X = \{q_1, q_2, q_3, \ldots\}$, we may set $\mu(\{q_i\}) = 2^{-i}b$ and then we'd have $\{\mu(E) : E \in \mathcal{S}\} = [0,b].$
Further, while this particular $X$ simplifies $\mathcal{S}$, there's really nothing special about it.  This method can be used to find a $\mu$ whose range is $[0,b]$ for any infinite $X$ when using the co-countable $\sigma$-algebra (or more generally any measure space with an infinite $\sigma$-algebra).

I'm not sure what one can say precisely about $\mu$ when it's not $\sigma$-finite.
