Problem 30 from Shakarchi Stein's book

Question

If $E$ and $F$ are measurable sets with $m(E)>0$ and $m(F)>0$. Prove that $E+F$ contains some interval.

I know that this problem is very popular in MSE and I found many topics but most of the solutions use Fourier transform, Lebesgue density theorem which I am not familiar yet.

I know the following two facts are true:

1. If $\alpha\in(0,1)$ and $m_*(E)>0$ then exists an open interval $I$ such that $m_*(E\cap I)\geq \alpha m_*(I)$.

2)If $E\subset \mathbb{R}$ with $m(E)>0$ then $E-E$ contains some neighborhood of zero.

My professor said that above problem could be solved via these problems.

I was thinking on this problem about a week but no results.

However, I have ideas: Let's $\alpha=\frac{9}{10}$ then exists open intervals $I$ and $J$ such that $m(E\cap I)\geq \frac{9}{10}m(I)$ and $m(F\cap J)\geq \frac{9}{10}m(J)$. I had idea to shift one of the intervals, say $I+a\subset J$ WLOG. But this did not give any good results. Anyway, the idea if shifting may not work if $I=(-\infty,a)$ and $J=(b,+\infty)$.

I would be very grateful if somebody will show how to solve this problems using my ideas. And please do not close this topic because other topic on this problem have quite advanced solutions and I would like to see more simpler using the ideas which I provide.

Answer 1

This solution was provided to me by my friend.

We show that the difference set $E-F$ contains an intervals under the same assumptions. This implies the desired conclusion because then the set $-F$ is also measurable and $m(-F)=m(F)>0$, so $E+F=E-(-F)$ contains interval.

Using fact 1) we know that there exists an open interval $I$ and an open interval $J$ such that $m(E\cap I)\geq \frac{3}{4}m(I)$ and $m(F\cap J)\geq \frac{3}{4}m(J)$. WLOG, also we may assume that $E$ and $F$ has finite measures and $m(I)\geq m(J)$. Then there exists $a\in \mathbb{R}$ such that $J+a\subseteq I$.

Write $\delta=\frac{1}{4}m(J)$. observe that for any $0<|c|<\delta$, the intervals $I$ and $J+a+c$ intersect in an interval $K$ of length more than $\frac{3}{4}m(J)$. Writing $E_0=E\cap K$ and $F_0=(F+a+c)\cap K$, we have $$m(E\cap I)=m(E_0)+m(E\cap(I-K))<m(E_0)+\frac{1}{4}m(J).$$

Question 1: I was thinking on an estimate $m(E\cap(I-K))<\frac{1}{4}m(J)$ but I was not able to understand how he get it. Can anyone explain it, please?

hence $$m(E_0)>\frac{3}{4}m(I)-\frac{1}{4}m(J)>\frac{3}{4}m(J)-\frac{1}{4}m(J)=\frac{1}{2}m(J).$$ And $$m(F\cap J)=m((F+a+c)\cap (J+a+c))=m(F_0)+m((F+a+c)\cap((J+a+c)-K))<m(F_0)+\frac{1}{4}m(J)$$ so $$m(F_0)>\frac{3}{4}m(J)-\frac{1}{4}m(J)=\frac{1}{2}m(J).$$ Note that $E_0\cup F_0$ is contained in $K$ and hence $m(E_0\cup F_0)\leq m(K)\leq m(J)$. But both $E_0$ and $F_0$ has measure strictly greater than $\frac{1}{2}m(J)$, so by the additivity of the measure we conclude that $E_0\cap F_0\neq \varnothing$. Thus $E\cap (F+a+c)\neq \varnothing$, so there exists $x\in F$ such that $x+a+c\in E$. Therefore $a+c\in E-F$. Since this is true whenever $0<|c|<\delta$, the interval $(a-\delta,a+\delta)$ is contained in $E-F$.

Question 2: It is obvious that $(a,a+\delta)$ and $(a-\delta,a)$ is contained in $E-F$. But does $a\in E-F$?

Would be very grateful for answers to my questions.

Answer 2

Here is a low-tech implementation of a more advanced idea. Although this may not be the simplest solution, I believe this reveals the key idea of the proof.

1. Proof modulo a technical ingredient

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We will assume the following statement.

Proposition. Let $E, F \subset \mathbb{R}$ be bounded and measurable. Define $f(x) = m(E \cap (x-F))$. Then

1. $f$ is continuous.
2. $\int_{\mathbb{R}} f(x) \, dx = m(E)m(F)$.

Now, without loss of generality, we may assume that $E$ and $F$ is bounded. Then $f(x) = m(E\cap(x-F))$ is continuous and not identically zero. So there is an open interval $I$ on which $f > 0$. But for each $x \in I$,

\begin{align*} m(E\cap(x-F)) > 0 &\quad \Rightarrow \quad \exists y \ : \ y \in E \text{ and } x-y \in F \\ &\quad \Rightarrow \quad x \in E + F. \end{align*}

Therefore $I \subseteq E + F$ and the claim follows.

2. Proof of the technical ingredient

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Lemma 1. If $E$ is a measurable subset of $\mathbb{R}$ such that $m(E) < \infty$, then for every $\epsilon > 0$, there exist disjoint open intervals $I_1, \cdots, I_n$ such that $m\left(E \triangle \cup_{i=1}^{n} I_i\right) < \epsilon$.

Proof. This is Theorem 3.4.(iv) of Chapter 1 in Stein & Shakarchi.

Lemma 2. Let $U = \cup_{i=1}^m (a_i, b_i)$ and $V = \cup_{j=1}^n (c_j, d_j)$ be disjoint unions of bounded open intervals. Define $f(x) = m(U \cap (x-V))$. Then

1. $f$ is continuous.
2. $\int_{\mathbb{R}} f(x) \, dx = m(U)m(V)$.

Proof. By writing $f(x) = \sum_{i,j} m((a_i, b_i) \cap (x-c_j, x-d_j))$, it suffices to check both claims when $U$ and $V$ are bounded open intervals, in which case those claims are easily verified by brutal force.

Lemma 3. For measurable subsets $A, B, C, D$ of $\mathbb{R}$ with finite measure,

1. $\lvert m(A) - m(B) \rvert \leq m(A \triangle B)$.
2. $\lvert m(A \cap B) - m(C \cap D) \rvert \leq m(A \triangle C) + m(B \triangle D)$.

Proof. 1 is a direct computation. For 2, use $(A\cap B) \triangle (C \cap D) = ((A\triangle C)\cap B) \triangle ((B \triangle D)\cap C)$.

Proposition. Let $E, F \subset \mathbb{R}$ be bounded and measurable. Define $f(x) = m(E \cap (x-F))$. Then

1. $f$ is continuous.
2. $\int_{\mathbb{R}} f(x) \, dx = m(E)m(F)$.

Proof. Let $M > 0$ be such that $E, F \subseteq [-M, M]$. For each $n$, use Lemma 1 to pick $U_n$ and $V_n$ satisfying:

• $U_n$ is a finite union of open intervals such that $U_n \subset [-M, M]$ and $m(E \triangle U_n) < 2^{-n}$.
• $V_n$ is a finite union of open intervals such that $V_n \subset [-M, M]$ and $m(F \triangle V_n) < 2^{-n}$.

Then $f_n(x) = m(U_n \cap (x-V_n))$ is continuous by Lemma 2 and $\lvert f(x) - f_n(x) \rvert \leq 2^{-(n-1)}$ by Lemma 3. So $f_n \to f$ uniformly on $\mathbb{R}$ and hence $f$ is continuous. Moreover, both $f$ and $f_n$ are supported on $[-2M, 2M]$. So

$$ \int_{\mathbb{R}} f(x) \, dx = \int_{-2M}^{2M} f(x) \, dx = \lim_{n\to\infty} \int_{-2M}^{2M} f_n(x) \, dx = \lim_{n\to\infty} m(U_n)m(V_n) = m(E)m(F). $$

3. Remarks

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After learning Lebesgue integration, Chapter 2 of Stein & Shakarchi, one can provide a much shorter proof of Proposition.

2^nd Proof of Proposition. Notice that $f(x) = \int_{\mathbb{R}} \mathbf{1}_E(y) \mathbf{1}_F(x-y) \, dy$. Then the continuity of $f$ follows the $L^1$-continuity of translation (Theorem 2.5 of Chapter 2). Then by Fubini's theorem,

$$\int_{\mathbb{R}} f(x) \, dx = \int_{\mathbb{R}^2} \mathbf{1}_E(y) \mathbf{1}_F(x-y) \, dxdy = m(E)m(F).$$