Dudley Section 3.4, Problem 1: Lebesgue Measure of Symmetric Difference is 0

Question

Struggling with Problem 1, section 3.4 of Dudley's Real Analysis and Probability. The question asks us to prove the following statement:

Let E be a Lebesgue measurable set, such that for all x in a dense subset of $\mathbb{R}$, $\lambda (E\Delta (E+x)) =0$. Show that either $\lambda (E)$ or $\lambda(\mathbb{R}/E) =0$.

My thoughts so far: If x is such that the two sets E and E+x are fully disjoint, then naturally their symmetric difference is simply the union of the two sets, implying $\lambda (E) =0$.

However I'm struggling to show what happens if they are not disjoint! Any help would be much appreciated.

Answer 1

Sets of positive measure on the real line have the following useful property: if $E\subset\mathbb{R}$ is of positive measure, then for any $\varepsilon>0$ there is an interval $I\subset\mathbb{R}$ such that $$m(E\cap I)>(1-\varepsilon)m(I).$$ That is to say, a set of positive measure can be arbitrarily "dense" on local regions. In your problem, suppose for the contrary that $E$ is of both positive measure and positive "co-measure", then you can find an interval $I$ such that $$m(E\cap I)>\frac{2}{3}m(I)$$ and an interval $J$ such that $$m(E^{c}\cap J)>\frac{2}{3}m(J)$$ which is equivalent to $$m(E\cap J)<\frac{1}{3}m(J).$$ By considering subintervals of $I$ and $J$, you can quite easily assume without loss of generality that $I$ and $J$ are of the same length. However, the condition "$m(E\Delta(E+x))=0$ for densely many $x$" says roughly that $E$ changes negligibly under densely many translations. But our result shows that $E$ behaves very differently on $I$ and on $J$, so a translation that moves $I$ approximately to $J$ will definitely deal a substantial change on $E$, i.e. yields a non-negligible symmetric difference.

Answer 2

Without loss of generality we may suppose the given dense set $D$ is countbale and symmetric about $0$. Also, we may assume that $E$ is symmetrc about $0$ (by looking at $E \cup -E$).

Let $F_0=E\setminus \bigcup_{x \in D} (E\Delta (E+x))$ and $F=F_0\cup -F_0$. Verify that $F+x=F$ for all $x \in D$. Note that $F$ is symmetric about $0$ and $\lambda (E \Delta F)=0$.

Now if $\lambda (E) >0$ and $\lambda (\mathbb R \setminus E) >0$ then $\lambda (F) >0$ and $\lambda (\mathbb R \setminus F) >0$. By a well known result in measure theory $F+(\mathbb R \setminus E)$ contains an open interval $(a,b)$. There is point $x$ of $D$ in $(a,b)$. But then $x=f+f'$ with $f \in F$ and $f'\in \mathbb R \setminus F$ contradicting the fact that $F+x=F$ for all $x \in D$.