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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?

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$P$ 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 $P(\bigcup_n A_n) =1$.

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