Prove that either $m(E)=0$ or $m(\mathbb{R}\setminus E)=0$.

Question

$\textbf{Problem :}$

Let $E\subset\mathbb{R}$ be a Lebesgue measurable set such that $m(E\setminus(E+x))=0$ for all $x\in\mathbb{R}$ ; where $E+x:=\{e+x:e\in E\}$.

Prove that either $m(E)=0$ or $m(\mathbb{R}\setminus E)=0$.

I have been able to prove it partially.

$\textbf{My Attempt :}$

Since $E$ is Lebesgue measurable, by definition, we have that, for any $A\subset\mathbb{R}$, $$ m^{\ast}(E)=m^{\ast}(E\cap A)+m^{\ast}(E\cap A^c). $$ In particular, we have that $m(E)=m(E\cap (E+x))+m(E\cap (E+x)^c)$ for all $x\in\mathbb{R}$.

Since $m(E\cap (E+x)^c)=m(E\setminus (E+x))=0$, we conclude that $m(E)=m(E\cap (E+x))$ for $x\in\mathbb{R}$.

Now notice that, for any $x\in\mathbb{R}$, $$m(E\cap (E+x))=\int_{\mathbb{R}}\chi_{E\cap (E+x)}(y)dy=\int_{\mathbb{R}}\chi_{E}(y)\cdot\chi_{(E+x)}(y)dy=\int_{\mathbb{R}}\chi_{E}(y)\cdot\chi_{E}(y-x)dy.$$ Hence, we have that, for any $x\in\mathbb{R}$, $$m(E\cap (E+x))=\int_{\mathbb{R}}\chi_{E}(y)\cdot\chi_{-E}(x-y)dy=(\chi_E\ast\chi_{-E})(x).$$ Since $m(E)=m(E\cap (E+x))$ for all $x\in\mathbb{R}$, we conclude that $$(\chi_E\ast\chi_{-E})(x)=m(E)\text{ for all }x\in\mathbb{R} $$

Now there are only two possibilities regarding the Lebesgue measure of $E$ : either $m(E)<\infty$ or $m(E)=\infty$.

Suppose that $m(E)<\infty$. This implies that $m(-E)<\infty$.

Therefore, we get that $\chi_{E}\in L^2(\mathbb{R})$ and $\chi_{-E}\in L^2(\mathbb{R})$. So, by a basic result in convolution, we conclude that $\chi_E\ast\chi_{-E}\in C_0(\mathbb{R})$. This means that we can make the function $\chi_E\ast\chi_{-E}$, which is the constant function $m(E)$, as arbitrary small as we would like outside a compact interval. This implies that $m(E)=0$.

Now I don't know how to proceed further when we have $m(E)=\infty$.

Please give me some hint. Thanks in advance.

Answer 1

This follows from Lebesgue's density theorem, the two following easily-confirmed facts for any $A \subset \mathbb{R}$:

• If $p \in A$ has density $1$ in $A$, then $p+x \in A+x$ has density $1$ in $A+x$.
• $A \setminus (A + x) = A \cap ((\mathbb{R}\setminus A) + x)$,

and the following lemma:

If $A_1,A_2 \subset \mathbb{R}$ and $x \in A_1 \cap A_2$ is a point of density $1$ in both $A_1$ and $A_2$ individually, then $x$ is a point of density $1$ in $A_1\cap A_2$.

To show the lemma, note that $x$ has density $0$ in both $\mathbb{R}\setminus A_1$ and $\mathbb{R}\setminus A_2$ individually, and note that $x$ has density $1$ in $A_1 \cap A_2$ if and only if it has density $0$ in $\mathbb{R} \setminus (A_1 \cap A_2) = A_1^c \cup A_2^c$.

From here, we let $\varepsilon > 0$ be arbitrary and use subadditivity of $m$ to show that for $B_\varepsilon = (x-\varepsilon, x+\varepsilon)$, we have $$\begin{align*}\frac{m((A_1^c\cup A_2^c)\cap B_\varepsilon)}{2\varepsilon} &= \frac{m((A_1^c\cap B_\varepsilon) \cup (A_2^c\cap B_\varepsilon))}{2\varepsilon} \\ &\leq \frac{m(A_1^c\cap B_\varepsilon)}{2\varepsilon} + \frac{m(A_2^c\cap B_\varepsilon)}{2\varepsilon} \xrightarrow{\varepsilon \to 0^+}0\end{align*}$$

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Now, let's see how your problem follows from these observations.

We assume $E \subset \mathbb{R}$ such that $m(E) > 0$ and $m(\mathbb{R}\setminus E) > 0$. By Lebesgue's density theorem, almost every point of $E$ has density $1$ in $E$, and similarly for $\mathbb{R}\setminus E$. From here, we are able to conclude that there is at least one point of density $1$ in each set, so we may let $x \in E$ and $y \in \mathbb{R}\setminus E$ have density $1$ in their respective sets.

Then $x = y + (x-y) \in \mathbb{R}\setminus E + (x-y)$ has density $1$ in $\mathbb{R} \setminus E + (x-y)$.

Since $x \in E \cap (\mathbb{R} \setminus E + (x-y))$ has density $1$ in both $E$ and $(\mathbb{R} \setminus E + (x-y))$ individually, the above lemma tells us $x$ has density $1$ in their intersection.

Therefore $x$ has density $1$ in $E \cap ((\mathbb{R}\setminus E) + (x-y)) = E \setminus (E + (x-y))$

In particular, we conclude $m(E \setminus (E + (x-y))) > 0$.