Proving a given assignation over a sigma algebra is a measure. Let $X$ be a no countable set and $M=\lbrace A \subset X |\mbox{A is countable or} A^{c} \: \mbox{is countable} \rbrace$. If I define the assignation $$\mu(A)=\begin{cases} 0 & \: if A \: \mbox{is countable} \\\ 1 & if A^{C} \: \mbox{is countable} \end{cases}$$ I claim this is not a measure since it seems to happen the additivity doesnt work here. For example: Let $A_{1},A_{2}\in X$ and both subsets are countable, so $A_{1} \cup A_{2}$ is countable. Then by one side we have that $\mu(A_{1} \cup A_{2})=1$, and by other side we have that $\mu(A_{1})+ \mu(A_{2})=1+1=2$ so $$\mu(A_{1} \cup A_{2}) \neq \mu(A_{1})+ \mu(A_{2}).$$

In fact the your set function is a measure in that sigma algebra.

Indeed let $\\{A_n:n \in \Bbb{N}\\}$ disjoint.

If all are countable then their countable union is countable so $$\mu(\bigcup_n A_n)=0=\sum_n\mu(A_n)$$

If some $A_m$ is uncountable then $A_n \subseteq A_m^c, \forall n \
eq m$ and by definition of the sigma algebra $A_m^c$ is countable.

So $\mu(A_n)=0 ,\forall n\
eq m$ and $\mu(A_m)=1$

Since $(\bigcup_nA_n)^c \subseteq A_m^c$ we have that $\mu(\bigcup_nA_n)=1=\sum_n\mu(A_n)$

If $\\{A_n\\}_{n \in \Bbb{N}}$ is any countable collection of pairwise disjoint sets in the sigma algebra, then it can contain at most one uncountable set because if $A_n,A_m$ are two disjoint uncountable sets, then $A^c_n,A^c_m$ are countable, so $A^c_n\cup A^c_m$ is countable, and thus $A_n \cap A_m=(A^c_n \cup A^c_m)^c$ is uncountable thus non-empty, which is a contradiction,since the sets are assumed disjoint.