I read in an old book the following example of a measure:
For the set $M=\mathbb{Q} \cap[0,1]$ denote with $S$ the set system of subsets of $M$ of the form $\mathbb{Q} \cap I$, where $I$ is any interval in $[0,1]$. Let us define the function $\mu: S \rightarrow \mathbb{R}$ as follows: for any set $A \in S$ of the form $A=\mathbb{Q} \cap I$ we set $\mu(A)=\ell(I)=b-a .$
Then it said without proof that $\mu$ is finitely additive, but not $\sigma$-additive.
As I did not get why I tried to prove it by myself and I tried to show that $S$ is a semi-ring, I guess that is important before I start with the other proof.
We have $\emptyset \in \mathbb{Q}$ and furthermore $(\mathbb{Q} \cap I_1)\cap (\mathbb{Q} \cap I_2)=\mathbb{Q} \cap I_1 \cap I_2$ and the union of two closed intervals is either an interval or the disjoint union of two intervals.
Then $(\mathbb{Q} \cap I_1)\setminus (\mathbb{Q} \cap I_2)$ is also an interval or the disjoint union of two intervals.
Now the proof. I do not quite understand how it cannot be $\sigma$-additive. Does it have something in common with Cantor sets? I don't know how to start the proof here. Any help or explanation (maybe an idea for the beginning of a proof) is appreciated. If there is a proof…
Best Answer
If $\mu$ is $\sigma$-additive then for disjoint suitable sets $A_n$ it must satisfy: $$\mu\left(\bigcup_{n=1}^\infty A_n\right)=\sum_{n=1}^{\infty}\mu(A_n)$$ From this it can be deduced that for suitable sets $B_n$ (not necessarily disjoint) it must satisfy:$$\mu\left(\bigcup_{n=1}^\infty B_n\right)\leq\sum_{n=1}^{\infty}\mu(B_n)$$
Now for every $r\in M$ choose an interval $I_r$ with $r\in S_r=\mathbb Q\cap I_r\in\mathcal S$.
If $\mu$ is indeed $\sigma$-additive then:$$1=\mu(M)=\mu\left(\bigcup_{r\in M}S_r\right)\leq\sum_{r\in M}\mu(S_r)=\sum_{r\in M}l(I_r)$$
However it is easy to choose the intervals $I_r$ in such a way that:$$\sum_{r\in M}l(I_r)<1$$and doing so we run into a contradiction.