Alternative proof : If E and F have positive measures then E+F contains an interval

Question

The statement is following :

$E, F$ are measurable subsets in $\mathbb{R}$ which have positive measure. Then prove that $E+F=\left\{ x+y\ :\ x\in E,\ y\in F\right\}$ contains an interval.

This is a problem 30 in Stein's Real analysis. I already read the proof of this by using convolusion which is here.(This problem is so-called Steinhaus theorem) Moreover, I proved that the previous problems number 28 and 29 in the book. My goal is applying the same technique to above one, using the technique in problem number 29 :

First, by problem number 28, $ \exists I^F$ and $I^E$ such that $$m(E\cap I^{E})\geq{\displaystyle \frac{9}{10}m(I^{E})} \text{ and } m(F\cap I^{F})\ge{\displaystyle \frac{9}{10}m(I^{F})}.$$ Let $E_{0}= E\cap I^{E}$ and $F_{0}= F\cap I^{F}$ and let $e_{0}, f_{0}$ : center of $I^{E}, I^{F}$, respectively.

Suppose $E_{0}+F_{0}$ does not contain an interval centered at $e_{0}+f_{0}$. Then $\exists a$ : small such that $E_{0}\cap(e_{0}+f_{0}+a-F_{0})=\phi$. But, $E_{0}\cup(e_{0}+f_{0}+a-F_{0})\subset I^{E}\cup(e_{0}+f_{0}+a-I^{F})$. Measure of left hand side is more than $\frac{9}{10}(m(I^{E})+m(I^{F}))$ but the measure of Right had side is slightly less than $\max(m(I^{E}),m(I^{F}))$.

And that's it. I stopped here. I can't make any contradiction. I thought that if I can make measure of $I^E and I^F$ are equal, then I prove the problem but this one is also a big problem.

Anyone can comment on this?

Answer 1

This proof uses the Lebesgue density theorem. If $e$ and $f$ are points of density of $E$ and $F$, and $\epsilon > 0$, there is $\delta > 0$ such that $m([e-\delta,e+\delta] \backslash E)/(2\delta) < \epsilon $ and $m([f-\delta,f+\delta] \backslash F)/(2 \delta) < \epsilon$.
I claim that $e + f + [-\delta/2,\delta/2] \subseteq E + F$.

Consider any $z \in e + f + [-\delta/2, \delta/2]$. Let $s$ be chosen from the interval $e + [-\delta/2, \delta/2]$. Thus $z-s \in f + [-\delta, \delta]$. The set of $s \in e + [-\delta/2,\delta/2]$ for which $s \notin E$ has measure $< 2 \delta \epsilon$. The set of $s \in e + [-\delta/2, \delta/2]$ for which $z-s \notin F$ has measure $< 2 \delta \epsilon$. If $4 \delta \epsilon < \delta$, removing these leaves a set of positive measure (and in particular at least one) in which $s \in E$ and $z-s \in F$, so that $z \in E + F$.

Answer 2

actually I think your idea could be extended by substituting $e+f$ to any real number, then you could move Ie until Ie is contained in another

Answer 3

First, $E + F$ can be replaced by $E - F$ by redefining $F$ as $-F$. (This will avoid some clustering later.)

Choose $I_1$ and $I_2$ such that

$$ \begin{eqnarray} m(I_1 \cap E) > (1-\epsilon) m(I_1)\tag{1} \\ m(I_2 \cap F) > (1-\epsilon) m(I_2)\tag{2} \end{eqnarray} $$

We can divide $I_1$ into consecutive sub-intervals of equal length. At least one of those sub-intervals will still satisfy (1). The same is true for $I_2$. Therefore, we can get one sub-interval $I'_1$ for $E$ and another $I'_2$ for $F$, such that $$(1+\epsilon) m(I'_2) \ge m(I'_1) \ge m(I'_2) \equiv \alpha.$$

In the text that follows, I will write the un-primed version.

By shifting $F$, we can make $I_1 \supset I_2$.

Let $A = I_1 \cap E$ and $B = I_2 \cap F$, and define $$ \begin{eqnarray} x &=& m(A-B) \\ y &=& m(A \cap B) \\ z &=& m(B-A) \end{eqnarray} $$

Taking the union of the three disjoint sets above, we have $$ \begin{eqnarray} x+y+z &\le& m(I_1) = (1+\epsilon) \alpha \tag{3} \\ (x+y) + (y+z) &>& (1-\epsilon) (m(I_1) + m(I_2)) = 2(1-\epsilon)\alpha \tag{4}. \end{eqnarray} $$

Subtracting (3) from (4), we have $y > (1-3\epsilon) \alpha$. By choosing e.g. $\epsilon = 0.1$ from the very beginning, we have $m(A\cap B)>0$.

By Exercise 29, the difference set of $A \cap B$, which is a subset of $E-F$, contains an open interval.