Sum of measurable sets

Kent Merryfield

If $A$ and $B$ are subsets of $\mathbb{R},$ define $A+B$ by $A+B=\{a+b:a\in A, b\in B\}.$ Suppose that $A$ and $B$ are measurable sets of positive finite measure. Prove that $A+B$ contains an open interval.

**Posted:** Tue Nov 09, 2004 1:17 am

Myth

Hi Kent!
Finally, I am at home!

Very nice problem.

Suppose contrary, i.e. $\frac{A+B}{2}$ doesn't contain intervals. Assume WLOG $A,B\subset [0,1]$ , then $\frac{A+B}{2}\subset [0,1]$ .
Let $n>1$ be natural number. Consider subintervals $\left(0,\frac{1}{n}\right)$ , ..., $\left(\frac{n-1}{n},1\right)$ . Each of these intervals contains point $c$ s.t. $c\not\in \frac{A+B}{2}$ . It follows that $(2c-A)\cap B=\varnothing$ .
So we have set $C\subset [0,1]$ s.t. $\overline{C}=[0,1]$ and $\forall\,c\in C:\ (2c-A)\cap B=\varnothing$ .
Since $B$ is a mesurable set, we can find open set $\hat{B}$ s.t. $B\subset \hat{B}$ and $|\hat{B}\setminus B|\leq \epsilon$ , where $\epsilon$ will be defined later. Any open set in $\mathbb{R}$ is a union of at most countable many pairwise disjoint intervals. So $\hat{B}=B_1\cup B_2\cup ...$ , where $B_k$ is an interval, $B_k\subset [0,1]$ .
For each interval $B_k$ we can find $c_k\in C$ s.t. $|(2c_k-A)\cap B_k|\geq \frac{1}{3}|B_k|\cdot |A|$ (!).
Thus we have
1) $\bigcup_{k=1}^\infty (2c_k-A)\cap B=\varnothing$ ;
2) $\left|\bigcup_{k=1}^\infty (2c_k-A)\cap \hat{B}\right|\geq \frac{1}{3}|\hat{B}|\cdot|A|$ ;
3) $|\hat{B}\setminus B|\leq \epsilon$ .
It follows that
$B''=\bigcup_{k=1}^\infty (2c_k-A)\cap \hat{B} \subset \hat{B}\setminus B=B',$
so $|B''|\leq |B'|$ . It yields
$\frac{1}{3}|\hat{B}|\cdot|A|\leq \epsilon.$
Take $\epsilon =\frac{1}{4}|B|\cdot|A|$ to obtain contradiction.

I think I also have another solution using characteristic functions and integrals.

**Posted:** Tue Nov 16, 2004 6:15 am

_________________
Myth is out of here

Kent Merryfield

**Posted:** Tue Nov 16, 2004 11:31 am

Myth

I am glad you will read it! Smile

P.S. I hope you checked my solution for topic "Convex function" http://www.artofproblemsolving.com/Forum/viewtopic.php?p=129665 Wink

**Posted:** Tue Nov 16, 2004 11:44 am

_________________
Myth is out of here

Kent Merryfield

**Posted:** Thu Nov 18, 2004 11:31 am

Myth

**Posted:** Thu Nov 18, 2004 12:01 pm

_________________
Myth is out of here

Kent Merryfield

**Posted:** Thu Nov 18, 2004 2:34 pm

Mindspa

A nice immediate corollary is that sets of positive measure contain hamel basis', the A - A contains a neighborhood of the origin analogue gives a proof of that any cantor set contain a hamel basis.

**Posted:** Sun Nov 21, 2004 7:25 am

_________________
Train hard, fight easy.

julien_santini

How do we jump from "any set with positive measure contains a Hamel basis" to "the Cantor set contains a Hamel basis" (since the Cantor set has measure $0$ ) ?

**Posted:** Tue Sep 26, 2006 5:09 pm

grobber

As has been discussed many times before on the forum (I remember seeing the problem at least three times; use the search function), if $C$ is the Cantor set, then $C+C=[0,2]$ (despite the fact that $C$ has measure zero; measure has nothing to do with this Smile

). This means that every real can be written as a rational multiple of a number in $C+C$ , which, in turn, means that $C$ contains a system of generators of $\mathbb R$ over $\mathbb Q$ . From any system of generators one can extract a basis, so that's that.

**Posted:** Tue Sep 26, 2006 9:41 pm

julien_santini

**Posted:** Tue Sep 26, 2006 10:13 pm

HilbertThm90

I was searching for posts on convolutions, and ran across this. I was excited to see that out of the two solutions given, the way I did this last week was not one of them.

There are $a_0$ and $b_0$ in $A$ and $B$ respectively with metric density 1. (Metric density at $x$ is $\lim_{r\to 0}\frac{m(E\cap B(x,r))}{m(B(x,r))}$ ). Since metric density is 1 at almost every point of a set, a sets of positive measure are guaranteed to have the points $a_0$ and $b_0$ as listed.

Choose $\delta>0$ to be small. Then let $c=a_0+b_0$ . For each $\epsilon$ , define $B_\epsilon=\{c+\epsilon-b: b\in B, \ |b-b_0|<\delta\}$ .

So we now have $B_\epsilon\subset(a_0+\epsilon-\delta, a_0+\epsilon+\delta)$ . So proper choosing of $\delta$ and $\epsilon$ small enough we have $A\cap B_\epsilon\neq\emptyset$ (by the metric density of 1), i.e. $A+B\supset (c-\epsilon_0, c+\epsilon_0)$ for some $\epsilon_0>0$ .

**Posted:** Thu Jul 31, 2008 5:05 pm

_________________
“Anyone who trades liberty for security deserves neither liberty nor security” - Ben Franklin