# A lower semicontinous but not sigma-additive measure

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 $$$$\mu((a,b]\cap\mathbb{Q}) = b-a .$$$$

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

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}.

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.