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My question is the following: let $m$ denote Lebesgue measure on $\mathbb{R}$. Define $A^2 := \{ x \cdot y \ | \ x, \ y \in A \}$. If $m(A) = 0$, show that $m (A^2) = 0$.

My initial observation was to try to write $$A^2 = \bigcup_{a \in A} a \cdot A$$ Then, certainly $m (a \cdot A) = 0$ for all $a$. However, $A$ may be uncountable, so we could have problems in the union. My attempt to fix this is to find an open set $E \supset A$ with measure $\varepsilon >0$, and consider any compact subset $K$ of $A^2$. We have that $\bigcup_{a \in A} a E$ is an open cover of $K$, whence we may choose a finite subcover $\{ a_1 E , \dots , a_n E \}$ of $K$. Then, $$m (K) < (a_1 + \dots + a_k ) \varepsilon$$ My problem with this is that, as $\varepsilon \to 0$, the number of sets needed to cover $K$ may become very large, so I cannot guarantee that $m (K) = 0$. I'd love to just say $m (A^2) = m(A)^2$, but that seems much too simple. Any help is appreciated!

Rellek
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1 Answers1

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This is false.

Let $B$ be the Cantor ternary set. This is an example of a set of null measure such that $B+B:=\{b+b' \mid b,b'\in B\}=[0,2]$ (cf this answer). Now let $A:=e^B=\{e^b \mid b\in B\}$. $A$ is of null measure since

$$m(A)=\int_\mathbb{R_+} 1_A(y) \,\mathrm{d}y=\int_\mathbb{R} 1_A(e^x) e^x \,\mathrm{d}x=\int_\mathbb{R} 1_B(x) e^x\,\mathrm{d}x=0, $$

while

$$A^2=\{aa' \mid a,a'\in A\} = \{e^be^{b'} \mid b,b'\in B\}=e^{B+B'}=e^{[0,2]}=[1,e^2] $$

has positive measure.

tristan
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  • Oh, wow. This was taken off a qualifying exam, I can't believe this was a mistake. Maybe they meant to make it a true/false question. Thanks! – Rellek Dec 23 '17 at 16:41
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    Perhaps they meant $A^2 = {x^2 \ | \ x \in A}$. That would've been true. – David Bowman Dec 23 '17 at 17:41