Show that $\mathbb P$ is finitly additive and that $\mathcal A$ is an algebra Given $\mathcal F=\{A \subset\Omega \mid A \mbox{ or } A^c \mbox{ is finite}\}$, show that $\mathcal F$ is an algebra.
Second, set:
$\mathbb P(A) = \begin{cases} 0, & \mbox{if } A \mbox{ is finite} \\ 1, & \mbox{if } A^c \mbox{ is finite} \end{cases}$
Show that $\mathbb P$ is finitely additive but not countably additive measure on $(\Omega,\mathcal A)$.
I have not taken any course in measure theory but still, the probability course of course requires that I know this vaguely. Anyone can help?
 A: Here is a similar situation. Consider $G=\{x\in\mathbb R\mid x^2+x=6\}$. How would you check that $\color{red}{\bf42}$ belongs to $G$ or that $\color{red}{\bf42}$ does not belong to $G$? You would have to determine whether $\color{red}{\bf42}$ satisfies the defining property of $G$ or not. To be in $G$, one must be a real number (and $\color{red}{\bf42}$ is) and one must solve the identity $x^2+x=6$ (and $\color{red}{\bf42}$ does not, since $\color{red}{\bf42}^2+\color{red}{\bf42}=1706\ne6$). Conclusion: $\color{red}{\bf42}$ is not in $G$. Likewise, is $\color{green}{\bf2}$ in $G$? Check: $\color{green}{\bf2}$ is a real number and $\color{green}{\bf2}$ solves $x^2+x=6$ (since $\color{green}{\bf2}^2+\color{green}{\bf2}=6$), hence $\color{green}{\bf2}$ is in $G$.
Back to your problem. Users in the comments suggested that you first check that $\color{purple}{\bf\Omega}$ belongs to $F=\{A\subseteq\Omega\mid A\ \text{or}\ A^c\ \text{is finite}\}$. How to check this? Just like before: to be in $F$, one must be a subset of $\Omega$ (so you have to determine whether $\color{purple}{\bf\Omega}$ is a subset of $\Omega$ or not) and one must be finite or have a finite complement (so you have to determine if at least one of the sets $\color{purple}{\bf\Omega}$ or $\color{purple}{\bf\Omega}^c$ is finite, or not).
Your conclusion? Is $\color{purple}{\bf\Omega}$ in $F$ or not?
