Any hints on how to show that this measure $\nu$ is finitely additive but not countably additive.

Question

Define the function $\nu: \mathcal M \to [0,\infty]$ by $\nu(A)=\infty$ if $0$ is in the closure of $A$ and $\nu(A) = 0$ otherwise. Prove that $\nu$ is finitely additive but not countably additive.

$\mathcal M$ is the set of Lebesgue measurable subsets of $\mathbb R$.

Any hints on how to show this?

Answer 1

Hint: consider a sequence of measurable sets $A_n$ such that $0 \notin \overline{A_n}$ for any $n$ but such that $0 \in \dots$

Answer 2

$\nu(\{1/n\})=0$ for each $n\in \mathbb N$ but $0\ne \nu(\{1/n:n\in \mathbb N\}).$

For finite additivity use the fact that $Cl(\bigcup_{i=1}^nA_i)=\bigcup_{i=1}^nCl(A_i)$ for any $n\in \mathbb N$.

Answer 3

for a fail of countable additivity take any infinite sequence $S$ consisting of distinct points in $\mathbb{R} \setminus \{0\}$ and for which $0 \in \bar S$