9
$\begingroup$

Let $A,B$ be subsets in $\mathbb{R}$. Is it true that $$m(A+B)=m(A)+m(B)?$$ Provided that the sum is measurable.

I think it should not be true, but could not find a counterexample.

$\endgroup$
4
  • 4
    $\begingroup$ What if $A=B=K$ where $K$ is a Cantor set? In such a case $\mu(K)=0$ but $K+K$ contains an interval. $\endgroup$ – Jack D'Aurizio Oct 17 '16 at 9:12
  • 1
    $\begingroup$ This might help you ! $\endgroup$ – Nizar Oct 17 '16 at 9:18
  • $\begingroup$ To summarize the answers, basically if $A$ and $B$ are intervals, then yes, there is equality, but if they are not connected, then the measure of the sum can be a lot larger. $\endgroup$ – Arthur Oct 17 '16 at 21:26
  • $\begingroup$ I am not sure that A+B is necessarily measurable. $\endgroup$ – DanielWainfleet Oct 18 '16 at 7:20
17
$\begingroup$

I preassume that here $A+B:=\{a+b\mid a\in A, b\in B\}$.

Counterexample (for Lebesgue measure):

Take $A=[0,1]\cup[2,3]$ and $B=[0,1]$ (so that $A+B=[0,4]$)

$\endgroup$
8
  • $\begingroup$ It seems that your Lebesque example actually does satisfy the condition since the sum is just $[0,2]$. $\endgroup$ – Tobias Kildetoft Oct 17 '16 at 8:56
  • $\begingroup$ @TobiasKildetoft Thank you. Repaired now. $\endgroup$ – drhab Oct 17 '16 at 9:15
  • 1
    $\begingroup$ The Lebesgue measure in invariant under translations so $A+\{b\}$ has the same measure as $A$, If $B\neq\varnothing$ then the measure of $A+B$ will not be smaller than the measure of $A$. $\endgroup$ – drhab Oct 17 '16 at 13:46
  • 1
    $\begingroup$ @MarioCarneiro The answer to that is "no". If $A=\varnothing$ then also $A+B=\varnothing$. So if secondly $m(B)>0$ then $m(A+B)=0<m(B)=m(A)+m(B)$. $\endgroup$ – drhab Oct 17 '16 at 18:34
  • 4
    $\begingroup$ However, if $m$ is the Lebesgue measure, $A$ and $B$ are nonempty, and $A+B$ is measurable, then $m(A + B) \ge m(A) + m(B)$. This is the one-dimensional case of the Brunn-Minkowski inequality. $\endgroup$ – Sasho Nikolov Oct 17 '16 at 19:07
27
$\begingroup$

Another example: $$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}. $$

$\endgroup$
6
$\begingroup$

Let we consider the ternary representation of some number in $(0,1)$: $$ x= 0.\overline{1022101}_3 $$ Exploiting $0=\frac{0+0}{2},1=\frac{0+2}{2},2=\frac{2+2}{2}$ digit by digit, we may write $x$ as the average between two numbers $a,b$ $$ x = 0.\overline{1022101}_3 = \frac{0.\overline{0022000}_3+0.\overline{2022202}_3}{2}=\frac{a+b}{2} $$ with the property that all their ternary digit belong to $\{0,2\}$. It follows that if $K$ is a Cantor set in $[0,1]$, $\mu(K)=0$, but $\mu(K+K)\geq 2$, since $K+K$ contains every point of the interval $(0,2)$.


This argument also has a discrete analogue in terms of Sidon sets or additive bases.
For instance, if $Q$ is the set of integer squares and $E=Q+Q$, $E$ has density zero in $\mathbb{N}$, i.e. $$ \lim_{n\to +\infty}\frac{\left|E\cap[1,n]\right|}{n}=0,$$ but every natural number belongs to $E+E$ by Lagrange's four-squares theorem.
If we take $C$ as the set of integer cubes, $$ \lim_{n\to +\infty}\frac{\left|C\cap[-n,n]\right|}{2n+1}=0, $$ but $$ \forall n\in\mathbb{Z},\qquad n\in \left(C+C+C+C+C\right).$$

$\endgroup$

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.