# Show that $P$ is finitely additive but not countable additive.

Consider the algebra $$\mathcal{A}$$ which consists of finite disjunct union of intervals in $$(0,1)$$. For $$A \in \mathcal{A}$$, let $$P(A) = 1$$ if there exists an $$\epsilon > 0$$, such that $$(1/2,1/2 + \epsilon] \subset A.$$ If this is not true then $$P(A) = 0$$.

Show that $$A$$ is finitely additive, but not countable additive.

For finitely additive we want to show that $$P(\cup_{n=1}^{N}A_n) = \sum_{n=1}^{N} P(A_n).$$ Because all the intervals are disjunct I was thinking there is only one interval that contains $$(1/2,1/2 + \epsilon]$$ so that $$P(\cup_{n=1}^{N}A_n) = 1$$. But why wouldn't this work for showing it is countable additive?

$P$ is finitely additive: If $A$ and $B$ are disjoint sets in $\mathcal{A}$, then at most one of them contains a set of the form $(1/2,1/2+\epsilon]$, so either $P(A)=0$ or $P(B)=0$. If $P(A)=1$, then $P(B)=0$ and $P(A\cup B)=1$. If both of $A$ and $B$ are measure 0, then $A\cup B$ does not contain the set of the form $(1/2, 1/2+\epsilon]$ so $P(A\cup B)=0$.
$P$ is not countably additive: consider $A_n = (\frac{1}{2}+\frac{1}{n+2}, \frac{1}{2}+\frac{1}{n+1}]$. You can check that $P(A_n)=0$ for each $n$ but $\bigcup_n A_n = (1/2,1]$ so $P(\bigcup_n A_n) =1$.