Show that $\mu((a,b)):=b-a$ finitely additive.

Question

This question is related to Show that $\mu$ is finitely additive and Measure on the set of rationals. However, the first post did not give convincing answer and the second post mostly focused on the $\sigma$-additivity.

The question arises from A. N. Shiryaev's "Probability" Page $138$ Problem 1. It states as follows:

Let $\Omega:=[0,1]\cap\mathbb{Q}$ be the set of rational points of $[0,1]$, and $\mathcal{A}$ the algebra of sets each of which is a finite sum of disjoint sets $A$ of one of the forms $(a,b)\cap\mathbb{Q}$, $[a,b)\cap\mathbb{Q}$, $(a,b]\cap\mathbb{Q}$, and $[a,b]\cap\mathbb{Q}$. And define $\mu(A):=b-a$. Show that $\mu(A)$ is finitely aditive but not countably additive.

Note that he does not require $a,b\in\mathbb{Q}$. For unknown reason, this question is quite confusing to me. I have two questions.

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Firstly, I know that $\mu$ is not $\sigma$-additive in this case. Enumerate $[0,1]\cap\mathbb{Q}=\{x_{1},x_{2},\dots\}$. Then, $$E:=\bigcup_{n=1}^{\infty}[x_{n},x_{n}]\cap\mathbb{Q}=\bigcup_{n=1}^{\infty}\{x_{n}\}=[0,1]\cap\mathbb{Q},$$ and the union is disjoint. Therefore, $\mu(E)=1$. But for each $[x_{n},x_{n}]\cap\mathbb{Q}$, $\mu([x_{n},x_{n}]\cap\mathbb{Q})=x_{n}-x_{n}=0$, so $\sum_{n=1}^{\infty}\mu([x_{n},x_{n}]\cap\mathbb{Q})=0$. Hence, $\mu$ is not $\sigma$-additive.

But, how to show that $\mu$ is finitely additive? I mean... if you have $([a,b]\cup [c,d])\cap \mathbb{Q}$ where $[a,b]$ and $[c,d]$ are disjoint, how do you compute $\mu(([a,b]\cup [c,d])\cap \mathbb{Q})$? This disjoint union cannot form a new interval, right? Is $\mu$ even defined in this case?

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Second, I know how to show that $\mathcal{A}$ is an algera. In fact, checking the closure under complement and under finite unions for those intervals is sufficient for us to extend the proof for any sets from $\mathcal{A}$. If $\Omega=[0,1]$, and those intervals are the usual real intervals (resp. rational intervals), then $\mathcal{A}$ can only be an algebra, since $[0,1]\cap\mathbb{Q}\notin\mathcal{A}$ because it has empty interior with respect to $[0,1]$.

But in this case, since every $[x_{n},x_{n}]\cap\mathbb{Q}=\{x_{n}\}\in\mathcal{A}$, it turned out that $\mathcal{A}$ contains every rational of $[0,1]\cap\mathbb{Q}$. This makes me wonder if $\mathcal{A}$ is actually a $\sigma$-algebra, since it is really close to $2^{[0,1]\cap\mathbb{Q}}$. But I cannot finish the proof since $\mathcal{A}$ only has finite union of those intervals, so it does not contain every subset of $\mathbb{Q}$. On the other hand, I cannot find a counterexample to show that $\mathcal{A}$ is not a $\sigma$-algebra.

I may be just confusing myself but I am trapped. I apologize in advance if the question is stupidly obvious. Thank you!

Answer 1

It seems to me they have worded the problem badly. They only define things like $\mu((a,b)\cap\Bbb Q)$ and although $\mathcal{A}$ contains things like $((a,b)\cup(c,d))\cap\Bbb Q$ they did not define $\mu$ on such sets. So, really they are asking: prove that $\mu$ can be extended to a premeasure on $\mathcal{A}$, ie that evaluations such as $\mu(((a,b)\cup(c,d))\cap\Bbb Q):=d-c+b-a$ are well defined. A priori it might not be since the same union can be written as a finite union of different subintervals.

They want to know if $\mu$ has a finitely additive extension to a map over all of $\mathcal{A}$. You have correctly shown that there is no countably additive extension of $\mu$. So! $\mu$ and $\mathcal{A}$ must somehow violate the Carathéodory extension theorem’s hypotheses.

The algebra is not a $\sigma$-algebra since it does not contain things like: $$\bigcup_{n\in\Bbb N}[0,a_n)\cap\Bbb Q=[0,e^{-1})\cap\Bbb Q$$

Where $a_\bullet$ is an increasing sequence of positive rationals tending to $e^{-1}$ from below. That the union is not in $\mathcal{A}$ (why?) is a problem since each unionand is drawn from $\mathcal{A}$.

^^That example rested on my initial interpretation that $a,b$ must be rational. If not, we can still have examples like: $$\bigcup_{n\ge1}(2^{-(n+1)},2^{-n})\cap\Bbb Q\notin\mathcal{A}$$