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

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 $m(A^2)=0$.
