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I have read an example on a probability book about measure theory. For the semiring $\mathcal{A}=\{(a,b]\cap\mathbb{Q}:a,b\in\mathbb{R},a\leq b\}$, define the set function $\mu:\mathcal{A}\rightarrow[0,\infty)$ be \begin{equation} \mu((a,b]\cap\mathbb{Q}) = b-a . \end{equation}

I can show that $\mu$ is lower and upper semicontinous, but how to proof that $\mu$ is not $\sigma$-additive?

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Let $\left(0,1\right]\cap\mathbb{Q}$ be presented as set $\left\{ q_{n}\mid n=1,2,\dots\right\} $

For every $n$ find $a_{n},b_{n}\in\left(0,1\right]$ with $a_{n}<q_{n}\leq b_{n}$.

Then $\left(0,1\right]\cap\mathbb{Q}=\bigcup_{n=1}^{\infty}\left(\left(a_{n},b_{n}\right]\cap\mathbb{Q}\right)$.

Now assume that there is $\sigma$-additivity.

If there is $\sigma$-additivity then there is also $\sigma$-subadditivity so that: $$1=\mu\left(\left(0,1\right]\cap\mathbb{Q}\right)\leq\sum_{n=1}^{\infty}\mu\left(\left(a_{n},b_{n}\right]\cap\mathbb{Q}\right)=\sum_{n=1}^{\infty}\left(b_{n}-a_{n}\right)$$

However we can take the terms $b_{n}-a_{n}$ as small as we want, so that we can arrive at a RHS with a value smaller than $1$.

This contradiction allows us to conclude that the assumption is false, i.e. that there is no $\sigma$-additivity.

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