Positive measure sets $A_{1}$, $A_{2} \subset \mathbb{R}$ such that $(\forall x \in A_{1})(\forall y \in A_{2}) x-y \notin \mathbb{Q}$

Question

Here's the problem in full, and what I've got so far:

Let $E= \lbrace (x, y) \in \mathbb{R}^2 | x-y \in \mathbb{Q} \rbrace$

1. Find $m(E)$, where $m$ is the Lebesgue measure on $\mathbb{R}^2$.
2. Do there exist measurable sets $A_{1}$, $A_{2} \subset \mathbb{R}$ with positive Lebesgue measure such that $(A_{1} \times A_{2}) \cap E = \emptyset$?

I think I've solved $1$, but I'm stuck on $2$. Here's my solution of $1$: $E = \bigsqcup_{q \in \mathbb{Q}} \lbrace (x, y) \in \mathbb{R}^2 | y = q+x \rbrace$, so $E$ is a countable union of lines, which are zero measure sets in $\mathbb{R}^2$, so $m(E)=0$.

I'm almost sure that the answer to $2$ is "no", but I can't get a contradiction by assuming the existence of such sets.

The only concrete thing I've got for $2$ is that for every $r_{1}$, $r_{2} \in \mathbb{Q}$, $(r_{1} + A_{1}) \cap (r_{2} + A_{2}) = \emptyset$, and if we put $r_{1} = r_{2} = 0$, we get that $A_{1} \cap A_{2} = \emptyset$. I know about Why can't there be a bounded set with positive Lebesgue measure such that $\forall x,y$ in it, $x-y\notin\mathbb Q$? , but I don't know how to use it here (or if I should even use it).

Answer 1

The answer is indeed no, by Lebesgue's density theorem.

Suppose $m(A)>0$ and $m(B)>0$ for $A,B \subset \mathbb{R}$. Then almost every $a \in A$ is a density point, and almost every $b \in B$ is a density point. For given density points $a \in A$ and $b \in B$ we can find a mutual $r>0$ such that at least 90 percent of $B_r(a)$ is in $A$ (in the sense that $\displaystyle \frac{m(A \cap B_r(A))}{m(B_r(a))}>.9)$, and at least 90 percent of $B_r(b)$ is in $B$. Then since $a$ is a density point of $A$, $-a$ is a density point of $-A$. Moreover, $-a+q$ is a density point of $-A+q$, for an arbitrary $q$. Now, from the density of the rationals, we can find such a $q$ that $|b-(-a+q)| < \displaystyle \frac{r}{100^{100}}$ (this is chosen for dramatic effect). And since $-a+q$ and $b$ were chosen so closely, we know that $B_r(-a+q) \cap B_r(b)$ is nonempty. Thus there are $x \in A$ and $y \in B$ such that $-x+q = y$, and this is what we wanted to show.

Answer 2

Incorrect answer, but I'm keeping this hoping that part of the arguments can be reused.

Assume such $A_1,A_2$ exist. Then they have positive Lebesgue measure. Similar to the proof in the linked question, define $Q=\Bbb Q\cap[0,1]$. Note that since all sets here is measurable, so measure and product are interchangeable. (measure of product of measurable sets = product of measures of measurable sets)

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As OP points out,$A_1,A_2$ can be unbounded, so the answer in the linked question can't be used directly.

As a result, we are going to show it suffices to consider the case for bounded $A_1,A_2$ using the standard truncation argument.

For $i = 1,2, n \in \Bbb{N}$, define $\bbox[2px, border: 1px solid black]{A_i^{(n)}=A_i \cap [-n,n]}$ so that it becomes bounded. Once we find some $n_1,n_2$ such that $A_1^{(n_1)}, A_2^{(n_2)}$ have positive Lebesgue measure, then the linked proof can be reused in this question.

By the continuity of measure, $n_i\uparrow\infty$, $m(A_i^{(n_i)})\uparrow m(A_i)>0$, so there exists $N_i$ such that $\bbox[2px, border: 1px solid black]{m(A_i^{(N_i)})>0}$.

You have assumed $(A_{1} \times A_{2}) \cap E = \varnothing$, so a fortiori $\bbox[2px, border: 1px solid black]{(A_{1}^{(N_1)} \times A_{2}^{(N_2)}) \cap E = \varnothing}$.

To simplify writing and to save ink, I'll strip off the superscript and denote $A_{1}^{(N_1)}$ as $A_1$ since these two unbounded given sets have no role to play in the rest of the proof.

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\begin{align} m((r_1+A_1)\times(r_2+A_2))&=m(A_1\times A_2) \forall r_1,r_2\in\Bbb{Q} \\ 0<m(Q+A_1)&\le b_1-a_1+1 \\ 0<m(Q+A_2)&\le b_2-a_2+1 \\ 0<m((Q+A_1)\times(Q+A_2))&=m(Q+A_1)\times m(Q+A_2)\\ &\le (b_1-a_1+1)(b_2-a_2+1) \end{align}

But \begin{align} &m((Q+A_1)\times(Q+A_2))\\ =&\sum_{r_1\in Q}\sum_{r_2\in Q} m(A_1+r_1)\times m(A_2+r_2)\\ =&\sum_{r_1\in Q}\sum_{r_2\in Q} m(A_1)\times m(A_2)\\ =&\infty, \end{align} which contradicts the above paragraph. Therefore, such $A_1,A_2$ don't exist.

Answer 3

This is not possible. It's a consequence of the Steinhaus theorem: if $A,B$ have positive measure, then $A-B$ has nonempty interior.

There is an analogous theorem for category, namely $A-B$ has nonempty interior if $A$ and $B$ are both non-meagre.