Measure of compact sets $C_{1}$ and $C_{2}$ given that $m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$ for any real number $x$.

Question

I'm trying to solve the following question in the context of another problem.

Suppose that $E$ is a Lebesgue measurable subset of $\mathbb{R}$ and let $C_{1}, C_{2}$ be compact subsets of $\mathbb{R}$ such that $C_{1} \subseteq E$, $C_{2} \subseteq E^{c}$ and $m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$ for any $x \in \mathbb{R}$ (where $m$ denotes the Lebesgue measure). Then, is it true that either $m(C_{1})=0$ or $m(C_{2})=0$?

I have been thinking about this for a while and I attempted to prove it by assuming that $m(C_{1}) >0$ and then trying to conclude that $m(C_{2})=0$ via the continuity of the real-valued function that maps every $x \in \mathbb{R}$ into $m(C_{1} \cap (C_{1}+x))$ and the fact that $m(C_{1} \cup (C_{2}+x)) = m(C_{2} \cup (C_{1}+x)) = m(C_{1})+m(C_{2})$ for every $x \in \mathbb{R}$ (which is implied by the condition $m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$ for any $x \in \mathbb{R}$). My intuition tells me that since we can guarantee that $C_{2}+x$ will intersect $C_{1}$ for an adequate real number $x$ and $m(C_{1}) >0$, then the fact the equalities $m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$ hold for any $x \in \mathbb{R}$ is a very strong condition that will imply that $m(C_{2})=0$ but I haven't really gotten anywhere besides those intuitive thoughts. Maybe the answer to the question is negative and there is a pathological counterxample with the fat Cantor set or something like that?

Any hints or ideas would be greatly appreciated. Thanks in advance.

Answer 1

Claim:$\;$If $A,B$ are Lebesgue measurable subsets of $\mathbb{R}$ such that for all $x\in\mathbb{R}$ we have $$ m\bigl(A\cap(B+x)\bigr)=0 $$ (where $m$ denotes Lebesgue measure), then $m(A)=0$ or $m(B)=0$.

Proof:

We'll prove the contrapositive.

Thus suppose $A,B$ are Lebesgue measurable subsets of $\mathbb{R}$ such that $m(A) > 0$ and $m(B) > 0$.

Our goal is to show that $ m\bigl(A\cap(B+x)\bigr) > 0 $ for some $x\in\mathbb{R}$.

By the Lebesgue Density Theorem

$\qquad$ Lebesgue density theorem in the line

there exists $a\in A$ and $b\in B$ such that the limits \begin{align*} & \lim_{r\to 0^+}\frac{m\bigl(A\cap(a-r,a+r)\bigr)}{2r}=1 \\[4pt] & \lim_{r\to 0^+}\frac{m\bigl(B\cap(b-r,b+r)\bigr)}{2r}=1 \\[4pt] \end{align*} both hold.

It follows that for some $r > 0$, the inequalities \begin{align*} & m\bigl(A\cap(a-r,a+r)\bigr) > r \\[4pt] & m\bigl(B\cap(b-r,b+r)\bigr) > r \\[4pt] \end{align*} both hold.

Now let $x=a-b$, and let $A_1,B_1$ be given by \begin{align*} & A_1=A\cap(a-r,a+r) \\[4pt] & B_1=\bigl(B\cap(b-r,b+r)\bigr)+x \\[4pt] \end{align*} We have $m(A_1) > r$.

By translation invariance $$m(B_1)=m\bigl(B\cap(b-r,b+r)\bigr)$$ so we get $m(B_1) > r$.

Clearly $A_1\subseteq (a-r,a+r)$, and also $$ B_1 = \bigl(B\cap(b-r,b+r)\bigr)+x \subseteq (b-r,b+r)+x $$ hence, since $(b-r,b+r)+x=(a-r,a+r)$, we have $B_1\subseteq (a-r,a+r)$.

Then $A_1\cup B_1\subseteq (a-r,a+r)$, so $m(A_1\cup B_1)\le 2r$.

Then we get $$ m(A_1\cap B_1) = m(A_1)+m(B_1)-m(A_1\cup B_1) > r+r-2r=0 $$ Then from $A\supseteq A_1$ and $B+x\supseteq B_1$, we get $$ m\bigl(A\cap(B+x)\bigr)\ge m(A_1\cap B_1) $$ hence $m\bigl(A\cap(B+x)\bigr) > 0$, which establishes the claim.

Answer 2

By assumption $$0=m(C_1\cap (C_2+x))=\int\mathbb{1}_{C_1}(y)\mathbb{1}_{-C_2}(x-y)\,dy=(\mathbb{1}_{C_1}*\mathbb{1}_{-C_2})(x)$$ As each $C_j$, $j=1,2$, is compact, $0\leq\mathbb(C_1),m(C_2)<\infty$. By Fubini's theorem and the translation and reflection invariance of the Lebesgue measure $$0=\int m(C_1\cap (C_2+x))=\int \mathbb{1}_{C_1}\mathbb{1}_{-C_2}(x)\,dx=m(C_1)m(-C_2)=m(C_1)m(C_2)$$

Hence either $m(C_1)=0$ or $m(C_2)=0$.