8
$\begingroup$

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, Lebesque 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.

$\endgroup$
  • 1
    $\begingroup$ There is no loss of generality in assuming that the sets have finite measure. ( Consider $E\cap (-n,n)$ and $F\cap (-n,n))$. In this case $I$ and $J$ are finite intervals. Try to complete your argument now. $\endgroup$ – Kavi Rama Murthy Sep 28 '18 at 23:23
  • $\begingroup$ @KaviRamaMurthy, I was trying in this way: There exists open intervals $I$ and $J$ such that $m(E\cap I)\geq \frac{9}{10}m(I)$ and $m(F\cap J)\geq \frac{9}{10}m(J)$. Suppose $m(I)\geq m(J)$ then $\exists x_0$ such that $J+x_0\subset I$. Let $E_0=E\cap I$ and $F_0=F\cap J$. Then $E_0+F_0\subset E+F$. I am trying to show that $E_0+F_0$ has open interval. Suppose it has not then for any $x\in E_0+F_0$ and $\forall \varepsilon>0$ we have $(x-\varepsilon,x+\varepsilon)$ is not subset of $E_0+F_0$. Unfortunately I don't know what to do further. $\endgroup$ – ZFR Sep 29 '18 at 4:25
  • $\begingroup$ @KaviRamaMurthy, The method which I applied for the problem $E-E$ is not applicable here. $\endgroup$ – ZFR Sep 29 '18 at 4:25
  • $\begingroup$ @KaviRamaMurthy, As you see something is wrong with my argument. Could you say how to prove it in a correct way? $\endgroup$ – ZFR Sep 29 '18 at 12:26
  • $\begingroup$ Sorry. I know several proofs of the statement but I do not know how to use 1) and 2) for this result. $\endgroup$ – Kavi Rama Murthy Sep 29 '18 at 12:29
2
$\begingroup$

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

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

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. $\lvert m(A) - m(B) \rvert = m(A \triangle B)$.
  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

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

2nd 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,

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ As I have not enough time for a detailed answer, let me be very brief: A1. You do not necessarily have $m(E\cap(I-K)) < \frac{1}{4}m(J)$. The main issue in the first property you listed is that, for each $\alpha$, the length of the so chosen interval $I$ cannot be controlled. So a priori the ratio $m(I)/m(J)$ can assume any value larger than $1$, say one billion. Then $m(E\cap(I-K))=m(E\cap I)-m(E\cap K)\geq \tfrac{3}{4}m(I)-m(J) = \left(\frac{3m(I)}{4m(J)}-1\right)m(J)$$ can become sufficiently large. $\endgroup$ – Sangchul Lee Oct 1 '18 at 0:16
  • $\begingroup$ A2. I see no reason to restrict $c$ to $0<|c|<\delta$. You may as well prove the statement for $|c|<\delta$ provide the rest of the argument can be corrected. Also, even if $c\neq 0$ turns out to be relevant for the proof, you still know that an interval $(a,a+\delta)$ is in $E-F$, which is enough. $\endgroup$ – Sangchul Lee Oct 1 '18 at 0:18
  • $\begingroup$ @SangchulLee, so you think the first estimate is wrong, right? $\endgroup$ – ZFR Oct 1 '18 at 0:21
  • $\begingroup$ I think so, at least in the given form. $\endgroup$ – Sangchul Lee Oct 1 '18 at 0:22
  • $\begingroup$ @SangchulLee, Thanks but do you know how to prove this problem using 1) and 2). $\endgroup$ – ZFR Oct 1 '18 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.