TITLE: A content which is lower and upper continuous, but not sigma additive
QUESTION [5 upvotes]: I'm trying to do this exercise:

I'm having trouble proving that $\mu$ is not $\sigma$-additive, since $\cup^{\infty}_{i=1}A_i=A\in \mathcal{A}$ I've been looking at sets $I_k$ with $\mu(I_k)=1/2^k$ or sets that give divergent series, but to no avail.
Any hint?
Edit: I thought I had found an answer in this page of Math.Stack, but the example there doesn't seem to work, since $\cup^{\infty}_{i=1}I_i\cap \mathbb{Q} \notin \mathcal{A}$ nor do they seem to be mutually disjoint...

REPLY [4 votes]: Since $\mathcal{A}$ is a semiring and $\mu$ is a content on $\mathcal{A}$, it follows that $\mu$ is a premeasure if and only if $\mu$ is $\sigma$-subadditive. Thus to find a counter-example for defying that $\mu$ is a premeasure, we only need to find a counter-example to show that $\mu$ is not $\sigma$-subadditive. To do this, as you mentioned, let $(x_k)_{k\in\mathbb{N}}$ be an enumeration of $(0,1]\cap\mathbb{Q}$ and let $B_k = (x_k-2^{-k-2},x_k+2^{-k-2}]\cap\mathbb{Q}\in\mathcal{A}$ for $k=1,2,\dots$.
Thus, $(0,1]\cap\mathbb{Q}\subset\bigcup_{k=1}^\infty B_k$, but $$\mu((0,1]\cap\mathbb{Q}) = 1 > \frac{1}{2}= \sum_{k=1}^\infty\mu(B_k)$$
This shows that $\mu$ is not $\sigma$-subadditive.
This exercise is very interesting..