Can we create a measure for infinitely countable sets with respect to eachother?

Consider the following function in the interval $[a,b]$ (between $x=a$ and $x=b$)

$$f(x)=\begin{cases}1 & x=A_1\\ 2 & x=A_2\setminus\left\{A_1\bigcap A_2\right\} \end{cases}$$

$$A_1=\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$$ $$A_2=\left\{\left.\sqrt{m}-\sqrt{n}\right|m,n\in \mathbb{Z}\right\}$$

(The domain is $x=\left\{{A_1}\bigcup{A_2}\right\}$ such that all possible points in both sets are between $[a,b]$)

I want to create an integral using a measure. In most measures for 1-d functions (lebesque, Haar measure, Dirac Measure), countable sets have a measure of zero but this is not useful for measuring countable sets with respect to eachother.

Both $A_1$ and $A_2$ appear to meet the requirements for a measure to exist but I'm not entirely sure. I know that $A_1$ and $A_2$ can be divided into infinitely disjoint unions.

So can we create a measure for $f(x)$ between $[a,b]$ that compares the measure for $A_1$ with respect to $A_2$ and vice versa? How could we apply such a measure for $f(x)$?

Currently Im not sure how to create a formal measure, so I decided to take the total sum of the number points that each of disjoint unions $\left(\text{on the right-side of the equalities} \, (1) \, \text{and} \, (2) \, \right)$ have between $[a,b]$.

I also believe that $2Q+1$ where $q$ is large and $\sqrt{N}$ where $n$ is large, we should set both values to $p$ inorder to also compare the density of $2q+1$ and $\sqrt{n}$. Hence for $2q+1=p$ we get $Q=\left\lfloor\frac{p-1}{2}\right \rfloor$ and $\sqrt{n}=p$ we have $N=\lfloor{p^2}\rfloor$.

In the case of $A_1=\left\{\left.\frac{2p+1}{2q+1}\right|m,n \in \mathbb{Z}\right\}$ we can set $A_1$ to equal $a$ and $b$ in terms of $p$. Then we can subtract the different $p$ values of $a$ and $b$ to get the number points $A_1$ has between $[a,b]$ for every $q$.

\begin{align} \frac{2p+1}{2q+1}=a&&\frac{2p+1}{2q+1}=b \end{align}

\begin{align} p=\frac{a(2p+1)-1}{2}&&p=\frac{b(2p+1)-1}{2} \end{align}

\begin{align} p=\left\lceil\frac{a(2p+1)-1}{2}\right\rceil&&p=\left\lfloor\frac{b(2q+1)-1}{2}\right\rfloor\end{align}

Substituting into $2p+1$ we should get the following sum.

$$\sum_{q=1}^{Q}\left(2\left\lfloor\frac{b(2q+1)-1}{2}\right\rfloor+1\right)-\left(2\left\lceil\frac{a(2p+1)-1}{2}\right\rceil+1\right)$$

$$(3)\quad D_{1}(p)=\sum_{q=1}^{\left\lfloor\frac{p-1}{2}\right\rfloor} 2\left\lfloor\frac{b(2q+1)-1}{2}\right\rfloor-2\left\lceil\frac{a(2p+1)-1}{2}\right\rceil$$

The same method can be done with $\left\{\left.\sqrt{m}-\sqrt{n}\right|m,n \in \mathbb{Z}\right\}$. Using $(2)$ we can solve in terms of $n$ for $b$ and $a$ and subtract both results to get the number of points $A_2$ has between $[a,b]$ for every $m$.

$$(4) \quad D_{2}(p)=\sum_{n=1}^{\left\lfloor p^2\right\rfloor} \left\lfloor{\left(b+\sqrt{n}\right)^{2}}\right\rfloor-\left\lceil{(a+\sqrt{n})^{2}}\right\rceil$$

It seems that if we take the following ratio we get $\lim_{p\to\infty}\frac{D_1(p)}{D_2(p)}=0$. Hence $D_2(q)$ is infinitely greater than $D_1(q)$. Hence $D_1(p)$ should have measure of $0$ and $D_2(p)$ should have a measure of $1$.

So by this new measure (integral), we would end up with the following.

$$0\int_{a}^{b}1+1\int_{a}^{b}2=\Bigl|_{a}^{b}x=b-a$$

Is the following the correct way of creating such a measure?

• Why is this tagged with the axiom of choice tag? – Asaf Karagila Nov 2 '16 at 20:24
• @Arbuja: what is $\land$ in $A_1\land{\left\{A_1\bigcap A_2\right\}}$? Do you mean the union of the two sets, which is simply $A_1$? Or the intersection, which is simply $A_1\cap A_2$? Or something else? – Greg Martin Nov 2 '16 at 21:12
• The axiom-of-choice tag would actually be quite appropriate, since it is used in a construction that gives a possible solution (see my answer). – Mikhail Katz Nov 3 '16 at 16:35
• The axiom of choice tag is actually inappropriate, because you're not asking about the necessity of the axiom of choice, you're asking about something whose existence may or may not require the axiom of choice. If this would be an appropriate criteria for this tag, then far more questions would fit into this tag. Questions tagged with the axiom of choice tag are generally questions which ask about specific interactions of the axiom of choice in proofs. Just because an answer appealed, indirectly, to the axiom of choice, does not make this question about the axiom of choice. – Asaf Karagila Nov 5 '16 at 20:14
• Have I said something is unclear? – Asaf Karagila Nov 6 '16 at 10:37

I find it a bit hard to follow all the details but perhaps you will find the following comment helpful. It is indeed possible to get measures that behave in a nontrivial fashion on countable subsets, and indeed the axiom of choice or some weaker forms thereof are relevant here. Namely, one can get a finitely-additive (not $\sigma$-additive) measure $\xi$ on the set of subsets of $\mathbb{N}$ such that $\xi(S)=0$ for any finite set $S\subseteq\mathbb{N}$ and $\xi(S)=1$ for any cofinite subset (i.e., subset with a finite complement). Here $\xi$ takes only two values, zero or one. For sets that are neither finite nor cofinite, the measure behaves in a nontrivial way but always exactly one of a pair of complementary sets has measure one. Such measures are used in ultraproduct constructions.
To apply this to the function stated in the post, one needs to include both sets in a single countable set. This can certainly be done. For example, one can take the extension $\bar {\mathbb Q}$ of $\mathbb Q$ consisting of all algebraic numbers; if you like you can just retain those that are real. This is a countable set and it obviously includes all the numbers involved in the sets you mentioned. However, which of the sets will have measure 1 and which will have measure 0 will depend on the choice of $\xi$.
• In general, in my post, I was creating a measure that sums the number of points set $A_1$ and $A_2$ can have between $[a,b]$. Then I compare the different sums and see which could have a greater measure. – Arbuja Nov 3 '16 at 16:24
• In your answer, could you show to apply ultraproduct constructions and axiom of choice with $f(x)$. I am currently taking Calculus and have insufficient knowledge to apply these concepts. – Arbuja Nov 3 '16 at 16:25
• Yes, so long as your countable sets are all subsets of a common countable set $X$. Then you just enumerate $X$ and use the measure $\xi$ to assign measures to your sets. – Mikhail Katz Nov 3 '16 at 16:29