$m(E\cap(E+t))=0,\,\,\,\forall t\neq0\implies m(E)=0$

Question

I want to prove that if $E$ is measurable and $m(E\cap(E+t))=0,\,\,\,\forall t\neq0\implies m(E)=0$. Here is my proof:

By the definition of Lebesgue measurability $$m(E)=m(E\cap(E+t))+m(E\cap(E+t)^c)=m(E\cap(E+t)^c)$$

$$\lim_{t\to0}m(E)=\lim_{t\to0}m(E\cap(E+t)^c)=m(E\cap E^c)=m(\emptyset)=0\implies m(E)=0$$

Is this proof fine? I am not really sure about the limit part. Thanks!

Yes, but why would $\lim_{t\to 0} m(E) = \lim_{t\to 0} m(E\cap (E+t)^c)$? — Michael L., Jul 30 '17 at 06:37
Makes sense so far. Now, why would $\lim_{t\to 0} m(E\cap (E+t)^c) = m(E\cap E^c)$? — Michael L., Jul 30 '17 at 06:41
@MichaelLee That part I am not too sure of. Can I say by the continuity of measure maybe? — Heisenberg, Jul 30 '17 at 06:42
There's the trick. What do you mean by "continuity of measure"? There are plenty of measures (such as any discrete measure) for which this does not hold. — Michael L., Jul 30 '17 at 06:43
Lebesgue measure in this case as I have mentioned in my proof as well — Heisenberg, Jul 30 '17 at 06:43
Yes, but can you formalize what "continuity of measure" means here? — Michael L., Jul 30 '17 at 06:44
Your question is a duplicate of this one, which also asks about continuity of measure. You might need to think more carefully about that step in your proof. — Michael L., Jul 30 '17 at 06:47
I went through it. I cannot really get if this is wrong or not though — Heisenberg, Jul 30 '17 at 06:55
The statement is true, but it's not really trivial. There's a proof here that $\lim_{t\to 0} m(E\cap (E+t)) = m(E)$, and from this you can conclude that as $m(E\cap (E+t)^c) = m(E)-m(E\cap (E+t))$, $$\lim_{t\to 0} m(E\cap (E+t)^c) = 0$$ — Michael L., Jul 30 '17 at 07:09
This is also a direct consequence of the Lebesgue density theorem. If you're interested I can post it as an answer — Bruno Andrades, Apr 27 '25 at 14:09

Michael L. · Answer 1 · 2025-04-27T13:43:10.310

To prove the problem statement, I'll use a strategy developed here.

First, let's handle the case where $m(E) < \infty$ for $E\subset \mathbb{R}^n$. Assume without loss of generality that $E$ is compact (as the Lebesgue measure is regular, we have that $m(E) = \sup\{m(F) : F\subseteq E, F\text{ compact}\}$; if $E$ is not compact, we can take some compact subset $F$ of $E$ and prove that $0 < m(F\cap (F+t))\leq m(E\cap (E+t))$). Then, we take an open set $U\supset E$ such that $m(U) < 2m(E)$. As $E$ is compact and $\partial U$ is closed, we have some minimal distance $\epsilon > 0$ between $E$ and $\partial U$ (see proof here). This implies that $E+x\subset U$ for all $x\in (-\epsilon, \epsilon)^n$. Therefore, $m(E\cap (E+x)) > 0$ for $x\in (-\epsilon, \epsilon)^n$, as otherwise, $$m(E\cup (E+x)) = m(E)+m(E+x)-m(E\cap (E+x)) = 2m(E) > m(U)$$ which contradicts that $E\cup (E+x)\subset U$.

In the case where $m(E) = \infty$, for any $\delta > 0$ we have some closed set $F\subset E$ such that $m(E\setminus F) < \delta$ (which implies that $m(F) = \infty$), and we will have some $r > 0$ such that $m(F\cap \overline{B_r(0)}) > 0$ for $\overline{B_r(0)}$ the closed ball of radius $r$ around $0$; otherwise, we would have $$m(F)\leq \sum_{r=1}^{\infty} m(F\cap \overline{B_r(0)}) = 0$$ Then, we can choose compact $E' = F\cap \overline{B_r(0)}$ such that $m(E') > 0$, and $m(E'\cap (E'+x)) > 0$ follows from above. This implies that if $m(E\cap (E+x)) = 0$ for all $x\neq 0$, we must have $m(E) = 0$.

I appreciate your answer. However, I already have an answer with me. The one I posted is one of my own ideas which I wanted to verify here. — Heisenberg, Jul 30 '17 at 06:41
Sorry for presuming! I think your line of thought is tempting, but the line $\lim_{t\to 0} m(E\cap (E+t)^c) = m(E\cap E^c)$ is not as trivial as you might assume. — Michael L., Jul 30 '17 at 07:10

$m(E\cap(E+t))=0,\,\,\,\forall t\neq0\implies m(E)=0$

1 Answers1