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

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.

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?

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!

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: $$\bigcup_{n\ge1}(2^{-(n+1)},2^{-n})\cap\Bbb Q\notin\mathcal{A}$$

• Great answer. I am sorry for the late reply. For the counterexample of $\mathcal{A}$ not being a $\sigma$-algebra, recall that the sequence $(1+\frac{1}{n})^{1+n}\searrow e$ as $n\rightarrow\infty$, and thus $\bigcup_{n=1}^{\infty}(0,(1+\frac{1}{n})^{-1-n})\cap\mathbb{Q}=(0,e^{-1})\cap\mathbb{Q}$. Therefore, I guess the problem in the book misses the requirement that for all those intervals $(a,b)\cap\mathbb{Q}$ and so on, $a,b\in\mathbb{Q}$ must be required. Otherwise there is just too much freedom. Apr 25, 2023 at 13:01
• For example, the counterexample you suggested, if without the requirement that $a,b\in\mathbb{Q}$, then $(0,e^{-1})\cap\mathbb{Q}$ still belongs to $\mathcal{A}$. Therefore, as you mentioned, the problem itself is quite poorly written. Apr 25, 2023 at 13:02
• I don't quite understand your answer about the measure though. Caratheodory says that if $\mu_{0}:\mathcal{A}\longrightarrow [0,\infty]$ is a countably additive set function on an algebra $\mathcal{A}$, then there exsists a measure $\mu$ on $\sigma(\mathcal{A})$ such that $\mu=\mu_{0}$ on $\mathcal{A}$. Yes, the $\mu_{0}$ in our example is not countably additive.. so what happens next? Apr 25, 2023 at 13:35
• @JacobsonRadical I just mentioned the Carathéodory extension theorem as a thing worth noting: here is an example of a finitely additive premeasure that can’t be extended. But it’s not relevant to the problem per se. I had just assumed they wanted $a,b\in\Bbb Q$, but if not then there are plenty of other examples. For instance: $$\bigcup_{n\in\Bbb N}(2^{-(n+1)},2^{-n})\cap\Bbb Q$$Is not in $\mathcal{A}$ despite every unionand being in $\mathcal{A}$. Apr 25, 2023 at 16:06
• Ok. I get it, and thanks for another counterexamples. For some reason this problem just made my brain stuck. Do you know how to show that it can be extended to a finitely additive measure? Any hint will be enough. Apr 25, 2023 at 21:56