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If $\mu: A \longrightarrow \left\{0,\infty\right\}$ is a set function where $A$ is the set of all left open right closed intervals in $(0,1]$ defined as: $$ \mu\bigl((a,b]\bigr) = \left\{ \begin{array}{l l} b-a & \quad \text{if $0\lt a\lt b\leq 1$ and $a\neq 0 $ }\\ +\infty & \quad \text{otherwise}\\ \end{array} \right.$$

then show that $\mu$ is not countably additive.

It is obviously finitely additive as I could show, but I'm not too sure how to show it will fail for countable additivity.

I tried writing $A = (0,1] = \bigcup_{i=1}^{\infty} ({1\over i+1},{1\over i}]$. Then $ \mu(A) =1$ but can I conclude that the RHS is not 1?

Could anyone help?

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Your example will work ... the right hand side is indeed 1 ... but $\mu(A) = \mu(0,1] = \infty$ by definition ... – martini Oct 1 '12 at 9:26
Did you mean $\mu\colon A\to\{0,\infty\}$ or to $[0,\infty]$? The former is just two possible values. – Asaf Karagila Oct 27 '12 at 9:44

We have $\mu(A)=\mu((0,1])=+\infty$ by definition, as $b=1$ and $a=0$ in the definition. But as $\frac 1{i+1}>0$, $$\sum_{i=1}^{+\infty}\mu\left(\left(\frac 1{i+1},\frac 1i\right]\right)=\sum_{i=1}^{+\infty}\left(\frac 1i-\frac 1{i+1}\right)=1.$$

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