Find the $\sigma$-algebra that makes the following map a measure($X=\mathbb{R}$) $v(A)=\begin{cases} 
      0 & \text{if $A$ is finite} \\
      1 & \text{if $A^c$ is finite} \\
   \end{cases} $
This is a basic question and I don't think $P(\mathbb{Q})$ is a desired $\sigma$-algebra, nor are $P(\mathbb{R})$ and $\{\emptyset, \mathbb{R}\}$. Any idea?
 A: Considering $X=\mathbb{R}$, you want to find the $\sigma$-algebras that make the following map a measure:
$\nu(A)=\begin{cases} 
      0 & \text{if $A$ is finite} \\
      1 & \text{if $A^c$ is finite} \\
   \end{cases} $
Note that there are more than one such $\sigma$-algebra.
$1$. Let $\Sigma$ be any $\sigma$-algebra generated by a finite collection of finite subsets of $\Bbb R$. That means there is $\{F_i\}_{i \in \{1, \dots, n\}}$ where each $F_i$ is finite and
$\Sigma = \sigma(\{F_i\}_{i \in \{1, \dots, n\}})$.
It is straight forward to check that $\nu$ is a measure defined on $\Sigma$.
$2$. Suppose $\Sigma$ is $\sigma$-algebra such that $\nu$ is measure defined on $\Sigma$.
First, if $E \in \Sigma$ is any infinite set, then $E^c$ is finite. In fact, if $E$ is a infinite set and $E^c$ is not finite, $\nu$ is not defined for $E$, which is a contradiction to the fact that $\nu$ is defined on $\Sigma$.
Let $\mathcal{F}=\{F \in \Sigma : F \textrm { is finite }\}$. From the previous paragraph, it follows immediately that $\Sigma = \sigma(\mathcal{F})$.
Second, let us prove that the set $\mathcal{F}$ is a finite collection.
Proof: Suppose  $\mathcal{F}$ is infinite and let $\mathcal{F}_1 \subseteq \mathcal{F}$ be a countably infinite subset of $\mathcal{F}$. Then $E=\bigcup_{F\in \mathcal{F}_1}F\in \Sigma$ is infinite. So, we know that $E^c$ is finite, and then we have
$$1= \nu(E) = \nu \left (\bigcup_{F\in \mathcal{F}_1}F \right) \leqslant \sum_{F\in \mathcal{F}_1} \nu(F)=0 $$
Contradiction.  $\square$
Combining $1$ and $2$ above we have that $\nu$ is a measure defined on the  $\sigma$-algebra $\Sigma$ if and only if $\Sigma$ is a $\sigma$-algebra generate by a finite collection of finite subsets of $\Bbb R$.
