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.
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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
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1$\begingroup$ This might help you ! $\endgroup$ – Nizar Oct 17 '16 at 9:18
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$\begingroup$ 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
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$\begingroup$ I am not sure that A+B is necessarily measurable. $\endgroup$ – DanielWainfleet Oct 18 '16 at 7:20
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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 $A+B=[0,4]$)
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$\begingroup$ 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
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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
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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
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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
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Another example: $$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}. $$
answered Oct 17 '16 at 9:49
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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
$$ \forall n\in\mathbb{Z},\qquad n\in \left(C+C+C+C+C\right).$$
answered Oct 17 '16 at 9:22
