Assume $\mu$ is the $U([0,1]^2)$ defined by
$$\mu([0,x]\times[0,y]=xy, \ (x,y) \in [0,1]^2$$
Show that $\mu$ assigns probability of $0$ to $\partial([0,1]^2)$
I tried to write each segment of the four segments of $\partial([0,1]^2)$ as a Cartesian product of two sets $[0,1]\times[0,1]$ to be able to use the definition?
Let $U=\{(x,1)\,|\,x\in [0,1]\}$, $R=\{(1,y)\,|\,y\in [0,1]\}$, $L=\{(0,y)\,|\,y\in [0,1]\}$, and $D=\{(x,0)\,|\,x\in [0,1]\}$. It is clear that $\partial ([0,1]^2)=L\cup R\cup U\cup D$.
Notice that for all $x,y\in [0,1)$ we have $$U\cup R\subseteq [0,1]^2\backslash [0,x]\times [0,y],$$ so $$\mu(U\cup R)\le \mu([0,1]^2)-\mu ([0,x]\times [0,y])=1-xy$$ for all $x,y\in [0,1)$, showing that $\mu(U\cup R)=0$. Since the Lebesgue measure is shift-invariant, we see that $\mu(U\cup R)=\mu(D\cup L)$. Thus, $$0\le \mu(\partial ([0,1]^2)=\mu(U\cup R\cup L\cup D)\le \mu(U\cup R)+ \mu(L\cup D)=0$$