 # I read in an old book the following example of a measure: For the set M = <mrow class="MJX lilmoore11p8 2022-07-14 Answered
I read in an old book the following example of a measure: For the set $M=\mathbb{Q}\cap \left[0,1\right]$ 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 $\mu :S\to \mathbb{R}$ as follows: for any set $A\in S$ of the form $A=\mathbb{Q}\cap I$ we set $\mu \left(A\right)=\ell \left(I\right)=b-a.$.
Then it said without proof that μ is finitely additive, but not σ-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 $\mathrm{\varnothing }\in \mathbb{Q}$ and furthermore $\left(\mathbb{Q}\cap {I}_{1}\right)\cap \left(\mathbb{Q}\cap {I}_{2}\right)=\mathbb{Q}\cap {I}_{1}\cap {I}_{2}$ and the union of two closed intervals is either an interval or the disjoint union of two intervals. Then $\left(\mathbb{Q}\cap {I}_{1}\right)\setminus \left(\mathbb{Q}\cap {I}_{2}\right)$ 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...
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Danika Rojas
If $\mu$ is $\sigma$-additive then for disjoint suitable sets ${A}_{n}$ it must satisfy:
$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_{n}\right)=\sum _{n=1}^{\mathrm{\infty }}\mu \left({A}_{n}\right)$
From this it can be deduced that for suitable sets ${B}_{n}$ (not necessarily disjoint) it must satisfy:
$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_{n}\right)\le \sum _{n=1}^{\mathrm{\infty }}\mu \left({B}_{n}\right)$
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:
$1=\mu \left(M\right)=\mu \left(\bigcup _{r\in M}{S}_{r}\right)\le \sum _{r\in M}\mu \left({S}_{r}\right)=\sum _{r\in M}l\left({I}_{r}\right)$

However it is easy to choose the intervals ${I}_{r}$ in such a way that:
$\sum _{r\in M}l\left({I}_{r}\right)<1$
and doing so we run into a contradiction.

We have step-by-step solutions for your answer!