Probability space Let $(X,\mathcal{M})$ be a $\sigma$-algebra where $\mathcal{M}:=\{A\subset X: A \hspace{0.2cm}\mbox{or} \hspace{0.2cm} X\setminus A \hspace{0.2cm}\mbox{is countable} \}$. I want to find all the $\mu\colon \mathcal{M} \longrightarrow [0,1]$ measures that turn $(X,\mathcal{M},\mu)$ into a probability space.
 A: For any probability measure $\mu$ we have that for any $A \in \mathcal{M}$ 
$$
\mu(A) = 1-\mu(X \setminus A).
$$
So $\mu$ is completely determined by its restriction to the set of countable sets.
Moreover if $A$ is infinite countable, then $A = \cup_{n =1}^\infty \{x_n\}$ for some sequence $(x_n)_n$ and $\mu(A) = \sum_{n=1}^\infty \mu(\{x_n\})$. For finite sets $A$, this sum will become finite of course. So $\mu$ is completely determined by the values it takes on the singletons $\{x\} \subset X$. Also note that $\mu(\{x\})$ can only be non-zero for a countable amount of $x$, otherwise $\mu$ can't be a finite measure. So this leads us to the definition of the following
$$
l^1(X) = \{ f: X \to \mathbb{C} \mid f \text{ is non-zero in a countable amount of points and } \|f\|_1 = \sum_{x \in X} \lvert f(x) \rvert <+\infty\}
$$
Of course this set is too large, but note that what I described above gives a bijection between all complex measures of $(X, \mathcal{M})$ and $l^1(X)$. Also note that $\sum_{x \in X} \lvert f(x) \rvert$ is well defined because $f$ is non-zero only in a countable amount of points. So really $\sum_{x \in X} \lvert f(x) \rvert = \sum_{n=1}^\infty \lvert f(x_n)\rvert$ for some sequence $(x_n)_n$ in $X$. Now we also have a bijection between the following set and all probability measures of $(X, \mathcal{M})$:
$$
S =\{ f \in l^1(X) \mid \|f\|_1 \leq 1 \text{ and } f(x) \geq 0 \text{ for all } x \in X\}
$$
To be more explicit, for any $f \in S$ you can define a measure 
$
\mu_f: \mathcal{M} \to [0,1]
$
by $\mu_f(A) = \sum_{ x \in A} f(x)$ for all countable $A$ and $\mu_f(A) = 1-\mu_f(X\setminus A)$ for all other $A \in \mathcal{M}$.
