# Additive but not $\sigma$-additive measure $\mathbb{Q} \cap I$

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...

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.
• Thanks, now I understand it better. What do you say about my "verification" of $S$ as a semi-ring? Oct 25, 2021 at 17:11
• Your proof of $\mathcal S$ being a semiring is a bit dubious because it is not clear to me what intervals are permitted. If e.g. they must be open then we are not dealing with a semiring. Note that e.g. $(0,1)-(0,0.5)=[0.5,1)$ cannot be written as union of open intervals. If open, closed and (half-open, half-closed) intervals are permitted then things are okay. Oct 25, 2021 at 17:45