I am considering the so called Cantor ternary set $C$ on $[0,1]$. I have just proved that its Lebesgue measure is $0$. To show that $C$ is nowhere dense, is it correct the following reasoning?
By contradiction, let's suppose there exists an open set contained in $C$. Without loss of generality, assume $\exists\,x_0\in[0,1]$ and $r>0$ such that the open ball $B(x_0,r)\subseteq C$. Then, by monotonicity of the Lebesgue measure $L^1$, we have that $L^1(B(x_0,r))=0$, which is an absurd.