Revisited question Lebesgue Measure of Symmetric Difference is 0

Question

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

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

I did not understand the first answer in this question. Can anyone add some details for that answer? Or another approach..

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Proof: I try to use the contradiction by following two results from For a set of positive measure there is an interval in which its density is high, $\mu(E\cap I)> \rho \mu(I)$:

Let $E$ be Lebesgue measurable, with $\mu(E)>0$ (here $\mu$ denotes the Lebesgue measure). Then:

1. for any $0<\rho<1$, there exists an open interval $I$ such that $\mu(E \cap I)> \rho \cdot \mu(I)$.
2. the set $E-E = \{x-y : x, y \in E\}$ contains an (open) interval centered at $0$ (in particular, if $\rho > 3/4$, the text I am using suggests that $(-\frac{1}{2} \mu(I), \frac{1}{2}\mu(I)) \subseteq E-E$).

Now, suppose that $m(E)>0$ and $m(E^c)>0$. Then for $\rho=\frac{3}{4}$ so that there is $I_1$ with $$ \mu(E\cap I_1)>\frac{3}{4}\mu(I_1) $$ and apply the result to $E^c$ to get there is a interval $I_2$ so that $$ \mu(E\cap I_2)<\frac{1}{4}\mu(I_2) $$

Then by the second result, we have there is an open interval so that $$ (-0.5m(I_1), 0.5m(I_2))\subset E-E $$ and $$ (-0.5m(I_2), 0.5m(I_2))\subset E^c-E^c. $$

Since $m(E\Delta (E+q))=0$ (i.e., $E=E^c$ a.e.), then for $E\cap (I_1\cup I_2)$ we have $$ E\cap (I_1\cup I_2)=(E+q)\cap (I_1\cup I_2) a.e. $$

Thus, $m(E\cap (I_1\cup I_2))=m((E+q)\cap (I_1\cup I_2))$

Then for $x\in (-0.5m(I_1), 0.5m(I_2))$, we have $x=a-b$ for some $a, b\in E$, and for $y\in (-0.5m(I_2), 0.5m(I_2))$, we have $y=c-d$ for some $c,d \in E^c$.

Answer 1

This is really just a restatement of @mathworker21 's answer, slightly simplified and clarified. I'm posting this per the OP's request as the OP remains confused about that answer.

As the OP already mentioned, if $\mu(E) > 0$ and $\mu(E^c) > 0$, then there is a finite open interval $I_1$ with $\mu(E \cap I_1) > \frac{3}{4}\mu(I_1)$ and a finite open interval $I_2$ with $\mu(E \cap I_2) < \frac{1}{4}\mu(I_2)$. Let $x_i$ be the center point of $I_i$ and $r_i$ be the radius of $I_i$, i.e., $I_i = (x_i - r_i, x_i + r_i)$ for $i = 1, 2$. The first point I'll observe here is that, by replacing $I_1$ or $I_2$ if necessary, I may assume $r_1$ and $r_2$ are close to each other. More precisely, I may assume $r_2 \leq r_1 < 3r_2$.

Indeed, if we have started with $r_1 < r_2$, then we may divide $I_2$ into finitely many intervals $\{J_k\}_{k = 1}^n$ with $J_k$ having radius $\frac{1}{2}r_1$ for all $k < n$ and $J_n$ having radius between $\frac{1}{2}r_1$ and $r_1$. (Just divide $I_2$ into equal intervals of radius $\frac{1}{2}r_1$. Then the remainder has radius smaller than $\frac{1}{2}r_1$. Add the remainder to the final interval to get an interval of radius between $\frac{1}{2}r_1$ and $r_1$.) Now, if all $J_k$ satisfies $\mu(E \cap J_k) \geq \frac{1}{4}\mu(J_k)$, then summing over all $k$ yields $\mu(E \cap I_2) \geq \frac{1}{4}\mu(I_2)$, contradicting the assumption on $I_2$. Hence, we may replace $I_2$ by a $J_k$ with $\mu(E \cap J_k) < \frac{1}{4}\mu(J_k)$. As $J_k$ has radius between $\frac{1}{2}r_1$ and $r_1$, we now have $\frac{1}{2}r_1 \leq r_2 \leq r_1$, i.e., $r_2 \leq r_1 \leq 2r_2 < 3r_2$.

Similarly, if we have started with $r_1 \geq 3r_2$, then we may divide $I_1$ into finitely many intervals $\{J_k\}_{k = 1}^n$ with $J_k$ having radius $r_2$ for all $k < n$ and $J_n$ having radius between $r_2$ and $2r_2$. By a similar argument, we may replace $I_1$ by one of the $J_k$. Then $r_2 \leq r_1 \leq 2r_2 < 3r_2$.

To summarize, there exists $I_i = (x_i - r_i, x_i + r_i)$ s.t. $\mu(E \cap I_1) > \frac{3}{4}\mu(I_1)$, $\mu(E \cap I_2) < \frac{1}{4}\mu(I_2)$, and $r_2 \leq r_1 < 3r_2$. Let $r = \frac{3r_2 - r_1}{4}$. Since $r_1 < 3r_2$, we have $r > 0$. Then the interval $(x_1 - x_2 - r, x_1 - x_2 + r)$ is open and nonempty, so there is $z \in (x_1 - x_2 - r, x_1 - x_2 + r)$ s.t. $\mu(E\,\Delta\,(E + z)) = 0$. Note that,

$$\begin{split} E\,\Delta\,(E + z) &\supseteq (E \cap I_1) \setminus (E + z)\\ &= (E \cap I_1) \setminus [(E + z) \cap I_1]\\ &= (E \cap I_1) \setminus ([E \cap (I_1 - z)] + z)\\ &= (E \cap I_1) \setminus ([E \cap (x_1 - z - r_1, x_1 - z + r_1)] + z) \end{split}$$

Let $s = \frac{3}{4}r_1 - \frac{1}{4}r_2 > 0$. Observe that, as $z \in (x_1 - x_2 - r, x_1 - x_2 + r)$, we have,

$$\begin{split} x_1 - z - r_1 \geq x_2 - r - r_1 = x_2 - \frac{3}{4}r_2 + \frac{1}{4}r_1 - r_1 = x_2 - \frac{3}{4}r_2 - \frac{3}{4}r_1 = x_2 - r_2 - s\\ x_1 - z + r_1 \leq x_2 + r + r_1 = x_2 + \frac{3}{4}r_2 - \frac{1}{4}r_1 + r_1 = x_2 + \frac{3}{4}r_2 + \frac{3}{4}r_1 = x_2 + r_2 + s \end{split}$$

Therefore,

$$\begin{split} (x_1 - z - r_1, x_1 - z + r_1) &\subseteq (x_2 - r_2 - s, x_2 + r_2 + s)\\ &= I_2 \cup (x_2 - r_2 - s, x_2 - r_2] \cup [x_2 + r_2, x_2 + r_2 + s) \end{split}$$

So,

$$\begin{split} \mu([E \cap (x_1 - z - r_1, x_1 - z + r_1)] + z) &= \mu(E \cap (x_1 - z - r_1, x_1 - z + r_1))\\ &\leq \mu(E \cap I_2) + \mu((x_2 - r_2 - s, x_2 - r_2]) + \mu([x_2 + r_2, x_2 + r_2 + s))\\ &< \frac{1}{4}\mu(I_2) + 2s\\ &= \frac{1}{2}r_2 + \frac{3}{2}r_1 - \frac{1}{2}r_2\\ &= \frac{3}{2}r_1 \end{split}$$

So,

$$\begin{split} \mu(E\,\Delta\,(E + z)) &\geq \mu((E \cap I_1) \setminus ([E \cap (x_1 - z - r_1, x_1 - z + r_1)] + z))\\ &\geq \mu(E \cap I_1) - \mu([E \cap (x_1 - z - r_1, x_1 - z + r_1)] + z)\\ &> \frac{3}{4}\mu(I_1) - \frac{3}{2}r_1\\ &= 0 \end{split}$$

This contradicts $\mu(E\,\Delta\,(E + z)) = 0$. So our original assumption must be wrong, i.e., we must have $\mu(E) = 0$ or $\mu(E^c) = 0$.

Answer 2

Assume both $m(E)$ and $m(\mathbb{R}\setminus E) =: m(E^c)$ are nonzero. Then each of $E$ and $E^c$ have positive Lebesgue measure. So there are points $x \in E$ and $y \in E^c$ for which $$\lim_{r \downarrow 0} \frac{1}{2r}m\bigl(E \cap (x-r,x+r)\bigr) = 1$$ and $$\lim_{r \downarrow 0} \frac{1}{2r}m\bigl(E^c \cap (y-r,y+r)\bigr) = 1.$$ Indeed, $m$-a.e. point of $E$ would work for $x$ and $m$-a.e. point of $E^c$ would work for $y$.

Note that the second equation is equivalent to $$\lim_{r \downarrow 0} \frac{1}{2r}m\bigl(E \cap (y-r,y+r)\bigr) = 0.$$ Now, take $r > 0$ small enough so that $$m\bigl(E \cap (x-r,x+r)\bigr) > \frac{3}{4}(2r)$$ and $$m\bigl(E \cap (y-r,y+r)\bigr) < \frac{1}{4}(2r).$$ By assumption, there is some $z \in (y-x-\frac{r}{10},y-x+\frac{r}{10})$ with $m\bigl(E \Delta (E+z)\bigr) = 0$. I'll let you finish the contradiction from here.