if $m(A)=0$ then $m(A^2)=0$ .$m$ is Lebesgue measure

Question

Let $A, B \subseteq \mathbb{R}$ and $m$ is Lebesgue measure on $\mathbb{R}$ now which of following options is true?

if A be uncountable then $m(A)>0$
if $m(A)>0$ then $A^{\circ} \neq \varnothing$
if $m(A)=0$ then $m(A^2)=0$ . (note that $A^{2}=\left\{a^{2} \mid a \in A\right\}$)
$m(A+B)=m(A)+m(B)$ (note that $A+B=\{a+b \mid a \in A, b \in B\}$ )

counterexample for 1 and 4 ,$A=B= $Cantor set and for 2 ,$[0,1] \cap \mathbb{Q}^{\mathrm{c}}$. how we can proof "3" ?

Just apply this with the function $f(x) = x^2$. – Arctic Char Aug 16 '20 at 20:13 — Arctic Char, Aug 16 '20 at 20:13

score 3 · Accepted Answer · answered Aug 16 '20 at 20:14

3

Compute\begin{align*} m(A^2) &= \int 1_{A^2}(x) \:dx \\ &= \int 1_{|A|}(\sqrt{x}) \: dx \\ &= \int_{|A|} 2u \: du \quad (\text{substitution with $u=\sqrt{x}$}) \end{align*}

and then note that $m(|A|) \leq m(A) + m(-A) = 2m(A) = 0$, so the above integral is over a set of measure $0$ and we conclude $m(A^2)=0$.

answered Aug 16 '20 at 20:14

Leander Tilsted Kristensen

5,488

What is $|A|$ ? – amir bahadory Aug 17 '20 at 13:52
I guess i should have added a definition, but i simply mean $|A|:={|x| : | : x\in A}$. – Leander Tilsted Kristensen Aug 17 '20 at 14:30

if $m(A)=0$ then $m(A^2)=0$ .$m$ is Lebesgue measure

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