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Problem: Let $m$ denote the Lebesgue measure on $\mathbb{R}$. Let $E$ be a Lebesgue measurable set of $\mathbb{R}$ such that $m(E \cap (E+t)) = 0$ for all $t \neq 0$. Prove that $m(E)=0$.

I am aware of several solutions to this problem, and some of them we can find here on Math.SE, but I am curious if I can also apply continuity of measure directly.

In particular, we know by the continuity of measure that if $\{B_k\}$ is an ascending sequence of measurable sets, then $m\left( \bigcup^{\infty}_{k=1} B_k\right)=\lim_{k\rightarrow \infty} m(B_k)$.

To specialize this for our case, suppose we consider the sequence $A_n = E \cap (E+\frac{1}{n})$ of sets. Clearly the conditions of the theorem are satisfied, since $m(A_n) < \infty$ and $A_1 \subseteq A_2 \subseteq \cdots$, and so we have that $m\left( \bigcup^{\infty}_{k=1} A_k\right)=\lim_{k\rightarrow \infty} m(A_k) = \lim_{k\rightarrow \infty} m(E \cap (E+\frac{1}{k})) = m(E)$, but this must be zero since unions of sets of measure zero is zero (left-most left-hand side of the equation).

Is this one way to go about this, and if not, where is the fault?

Thank you.

user125088
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  • Why do you conclude that the sets $A_n$ are nested? –  Jan 31 '14 at 00:20
  • @T.Bongers Ah, sorry, I meant the opposite inclusions, but then of course I will have unions instead. Wouldn't that work? – user125088 Jan 31 '14 at 00:36
  • Actually, the other inclusions are not true in general either. – i like xkcd Jan 31 '14 at 01:26
  • @Guillermo Ah, I was thinking of intervals I suppose. Could we salvage this proof if I approximate our measurable sets by open intervals? – user125088 Jan 31 '14 at 01:33
  • @Guillermo Also, do you have a counterexample in mind for when it fails in general? – user125088 Jan 31 '14 at 02:02
  • Maybe approximating will work. The example I had in mind was something like $E = {\frac{1}{2^k}:, k \in \mathbb{N}}$. It seems like $E \cap (E+\frac{1}{n})$ is oscillating between the empty set and a couple of numbers. – i like xkcd Jan 31 '14 at 15:08

1 Answers1

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We shall be using the following result:

Lemma. if $E$ is Lebesgue measurable and $m(E)>0$, then for every $\alpha\in (0, 1)$, there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.

Proof. It is a consequence of the Lebesgue Density Theorem, according to which, if $m(E)>0$, then for almost every point $x \in E$ $$ \lim_{\varepsilon\to 0} \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} = 1. $$ Hence for such an $x$ we can choose an $\varepsilon$, such that $$ \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} > \alpha $$ or equivalently $$ m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)> \alpha\,m\big((x-\varepsilon,x+\varepsilon)\big)\big). \tag*{$\Box$} $$

In our case, as $m(E)>0$, let an interval $I=[a,b]$, such that $m(E\cap I)>\frac{4}{5}m(I)$, and set $t=m(I)/5>0$. Clearly, writing $E_t=E+t$, $$ m\big(E_t\cap [a+t,b]\big)=m\big(E\cap [a,b-t]\big) \ge m(E\cap I)-m\big([b-t,b]\big)\ge \tfrac{3}{5}m(I), $$ and similarly $$ m\big(E\cap [a+t,b-t]\big),\,m\big(E_t\cap [a+t,b-t]\big)\ge\frac{2}{5}m(I), $$ while $m\big([a+t,b-t]\big)=\frac{3}{5}m(I)$. Therefore $$ m(E\cap E_t)\ge m\big(E\cap E_t\cap [a+t,b-t]\big)\ge\frac{1}{5}m(I). $$

Yiorgos S. Smyrlis
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