A content which is lower and upper continuous, but not sigma additive

Question

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...

Answer 1

Let $(x_k)_{k \in \mathbb{N}}$ be an enumeration of $\mathbb{Q} \cap (0,1)$. Define iteratively a sequence $(A_n)_{n \in \mathbb{N}} \subseteq \mathcal{A}$ as follows:

• $n=1$: $$A_1 := \mathbb{Q} \cap \bigg( \max\left\{0,x_1- \frac{1}{2^3}\right\}, x_1 \bigg].$$
• $n \to n+1$: Suppose that we have already defined $A_1,\ldots,A_n$. Set$$k_{n+1} := \inf\{j \geq 1; x_j \notin A_1 \cup \ldots \cup A_n\}$$ and define $$A_{n+1} := \mathbb{Q} \cap (x_{k_{n+1}}-r_{n+1},x_{k_{n+1}}]$$ where $r_{n+1} \in (0,2^{-n-3}]$ is chosen such that $A_{n+1} \subseteq (0,1)$ and $$A_{n+1} \cap (A_1 \cup \ldots \cup A_n) = \emptyset.$$

The so-defined sets are mutually disjoint and $\mathbb{Q} \cap (0,1) = \bigcup_{k \geq 1} A_k$. Since

$$\mu(\mathbb{Q} \cap (0,1)) = 1$$

but

$$\sum_{n \geq 1} \mu(A_n) \leq \sum_{n \geq 1} 2^{-n-2} = \frac{1}{4} < 1$$

we conclude that $\mu$ is not $\sigma$-additive.

Answer 2

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..