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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?

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  • $\begingroup$ Should I use the complement? $\endgroup$ – Note Mar 2 '18 at 23:38
  • $\begingroup$ can I apply $ \mu $ on a clopen interval, i.e. $[0,1)$ $\endgroup$ – Note Mar 3 '18 at 0:22
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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, $$0\le \mu(\partial ([0,1]^2)=\mu(U\cup R\cup L\cup D)\le \mu(U\cup R)+ \mu(L\cup D)=0$$

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  • $\begingroup$ I kind of arrived to a result similar where I derived $1-xy$. I guess $1-xy$ would equal $0$ because $x, y \in [0,1]$ $\endgroup$ – Note Mar 3 '18 at 2:10

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