Set in $\mathbb{R}^2$ that contains no (non-null) measurable rectangles

Question

From Torchinsky's Real Variables text, presented in a chapter on abstract Fubini's Theorem.

Define the following set: $$E := \mathbb{R}^2 \setminus \{(x,y) \in \mathbb{R}^2: x-y \in \mathbb{Q}\}.$$

Prove that $E$ does not contain any set of the form $A \times B$ where $A, B$ are Lebesgue-measurable, and $|A|,|B|>0$.

Answer 1

I am not quite sure where Fubini factors in, but your question can be answered by a slightly stronger statement.

Proposition Let $A, B\subset \mathbb{R}$ be Lebesgue measurable sets of positive measure, then the difference set $A - B$ contains an interval.

To wit, if $A - B$ contains an interval, it must contain a rational number, which would contradict that $A\times B\subset E$.

---

The proof of the proposition is a little bit involved. First we use the structure of the real line and the regularity of the Lebesgue measure to obtain

Lemma 1: If $A\subset\mathbb{R}$ is a measurable set with positive Lebesgue measure, then for every $\alpha \in (0,1)$ there exists an open interval $I$ such that $|A\cap I| \geq \alpha |I|$. That is to say, $I$ is almost entirely included in $A$.

Proof: without loss of generality we can take $A$ to be bounded and hence have finite measure. By the regularity of the Lebesgue measure there exists some open set $U\supset A$ such that $\alpha |U| \leq |A|$. Now, every open subset of the real line can be written as a countable disjoint union of open intervals. So writing $U = \sqcup U_{\beta}$ and $A = \sqcup (U_\beta \cap A)$ we have from pigeonhole principle at least one of the $U_\beta$ (which we take to be the interval $I$) must satisfy $\alpha |U_\beta| \leq |A\cap U_\beta|$.

Lemma 2: Let $A, B\subset\mathbb{R}$ have positive Lebesgue measure. For any $\alpha \in (0,1)$ we can choose open intervals $I_A$ and $I_B$ such that $|A \cap I_A| \geq \alpha |I_A|$, $|B\cap I_B| \geq \alpha |I_B|$ and $ |I_B| < |I_A| \leq 2 |I_B|$.

Proof: Observe that if $I_A = (x_1, x_2)$ is given by Lemma 1 for the set $A$, then at least one of $(x_1, \frac12(x_1+x_2))$ and $(\frac12(x_1+x_2), x_2)$ is also an interval satisfying the condition of Lemma 1.

So suppose via Lemma 1 we found intervals $I_A$ and $I_B$ such that $|I_A| > 2 |I_B|$, we can find some $k$ such that $2^{k+1} |I_B| < |I_A| \leq 2^{k+2} |I_A|$. So dividing $I_A$ in half $k$ times we can find a new $I_A$ satisfying the third condition. The situation with $|I_B| > |I_A|$ is similar.

---

Proof of proposition: Fix $\alpha > 15/16$. Let $I_A$ and $I_B$ be the intervals guaranteed to exist by Lemma 2. Let $A' = A\cap I_A$ and $B' = B \cap I_B$. Let $x_A$ be the midpoint of $I_A$ and $x_B$ be the midpoint of $I_B$, and let $L_A$ be the length of $I_A$ and $L_B$ the length of $I_B$. So in particular we have $$ L_B < L_A \leq 2 L_B $$

Since $|A'| > \frac{15}{16} I_A$ there exists $y_A \in A'$ such that $|y_A - x_A| < \frac1{16}$. Similarly we have $y_B$.

Let $z_0 = y_A - y_B$. We will show that $A' - B'$ contains a neighbourhood of $z_0$.

We proceed by contradiction. Suppose not, then for every $\epsilon \in (0, \frac12 (L_A - L_B))$ there exists $z_\epsilon \not\in A' - B'$ with $|z_\epsilon - z_0| < \epsilon$. This implies that the sets $B' + z_\epsilon$ is disjoint from $A'$. Thus

$$ | (B' + z_\epsilon) \cup A' | = |B'| + |A'| \geq \alpha ( L_A + L_B) \geq \frac32 \alpha L_A $$

On the other hand, we know that

$$ I_B + z_\epsilon \subset \left( - \frac9{16} L_B + y_B + z_\epsilon, \frac{9}{16} L_B + y_B + z_\epsilon \right) \subset \left( - \frac{9}{16} L_B + y_B + z_0 - \epsilon, \frac{9}{16} + y_B + z_0 + \epsilon \right) $$

so

$$ I_B + z_\epsilon \subset \left( - (\frac{9}{16} L_B + \epsilon) + y_A, (\frac{9}{16} L_B + \epsilon) + y_A\right) $$

On the other hand, we have

$$ I_A \subset \left( - \frac{9}{16} L_A + y_A, y_A + \frac{9}{16}L_A\right)$$

Thus, by our choice of $\epsilon$ we have that both $I_A$ and $I_B$ are contained in the interval

$$\left( - \frac{9}{16} L_A + y_A, y_A + \frac{9}{16} L_A\right) $$

so that

$$ \left| (I_B + z_\epsilon ) \cup I_A \right| \leq \frac{18}{16} L_A $$

But then we have

$$ \frac{45}{32} L_A \leq | (B' + z_\epsilon) \cup A' | \leq |(I_B + z_\epsilon) \cup I_A | \leq \frac{36}{32} L_A $$

which is absurd as $L_A > 0$.