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Let $E$ be a Lebesgue measurable subset of $\mathbb{R}$ such that $m(E\cap(E+t)) = 0$ for all $t \neq 0$. Then prove that $m(E)=0$.

I think that the function $f(t) = m(E\cap(E+t))$ is continuous but how could I used this result to prove $m(E) =0$?

Giovanni
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user145993
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1 Answers1

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Assume first that $E$ is bounded, then $f(t) = \int_E\chi_{E + t}(x)\, dx$ is continuous at $0$ by Lebesgue dominated convergence theorem. Then $$0 = \lim_{t \to 0^+}f(t) = f(0) = m(E).$$

If $E$ is not bounded we can apply the previous reasoning to $E_n = E \cap B(0,n)$, indeed $m(E_n \cap (E_n + t)) \le m(E \cap (E + t)) = 0$. This shows that $E = \bigcup E_n$ is the countable union of null sets, hence it has zero measure.

Alternatively, consider $E_N := E \cap B(0,N)$ and assume by contradiction that $m(E_N) > 0$. Notice that by construction $E_N + n^{-1} \subset B(0,N + 1)$ for every $n$ and that $$m(E_N \cap (E_N + n^{-1})) \subset m(E \cap (E + n^{-1})) = 0.$$

To obtain a contradiction notice that $$\infty > m(B(0,N+1)) \ge \bigcup_{n = 1}^{\infty}m(E_N + n^{-1}) = \sum_{n = 1}^{\infty}m(E_N) = \infty.$$

This second argument seems a little bit more powerful since by replacing $m$ with the outer measure $m^*$ we don't even need that $E$ is a measurable set.

Giovanni
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  • In the second argument. Why $\geq$ is true? – Qinsheng Zhang Feb 18 '21 at 04:26
  • @QinshengZhang: It follows from the fact that $(E_N + a) \cap (E_N + b) \subset E_N \cap (E + b - a)$. Therefore the intersections of the sets $E_N + n^{-1}$ have zero measure. Let me know if this helps, I can clarify further if needed. :) – Giovanni Feb 18 '21 at 11:55
  • I agree that $\cup(E_N + n^{-1} \in B$, however, it is not a direct result that $m(B) \geq \sum m(E_N + n^{-1})$. $\sum m(E_N + n^{-1}) \leq m(\cup E_N + n^{(-1)})$ – Qinsheng Zhang Feb 20 '21 at 02:56
  • We know that $m(B) \leq m(\cup E_N + n^{-1})$ but how can we conclude that $m(\cup E_N + n^{-1}) = \sum m(E_N + n^{-1})$, they are not disjoint. – Qinsheng Zhang Feb 20 '21 at 03:03
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    @QinshengZhang It is almost direct: you can use the inclusion-exclusion principle. (see for example this question: https://math.stackexchange.com/questions/2874519/proving-the-inclusion-exclusion-principle-for-measures). To make this precise we need one further step: we can take M large enough so that $M m(E_N) > m(B_N)$. Then we can restrict our attention to only a finite number of sets $E_N + t_m$, $m = 1, ..., M$. – Giovanni Feb 20 '21 at 18:05