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I tried to solve this Dividing open domains in $\mathbb R^2$ in parts of equal area

I want to know if my solution to (i) is "rigorous" and, since I proved a part of it in a different way, if it is correct.

Let $x,y, \in \mathbb{R}^2$; with $x-y$ I mean the direction of the line containing the two points. Let $A \subset \mathbb{R}$, with $\lvert A\rvert$ I mean the Lebsegue-$\mathbb{R}^1$ measure of $A$.

$(x_1, x_2):= \underset{(x,y) \in \mathbb{A^*}}{\text{argsup }} \|x-y\| \quad , \quad A^*:=\{(x,y) \in \bar{A}\times\bar{A} : (x-y,d)=0\} \quad , \quad l:=\|x_1-x_2\|$ Note $x_1, x_2, l$ are well defined since $A$ is bounded.

Let $r_{x_i}$ be the line of direction $d$ and containing $x_i$. I fix an orthogonal system $(xOy)$ where $x=r_{x_1}$ arbitrarily oriented, $y$ is a line perpendicular to $x$ oriented from $x$ to $r_{x_2}$ and $O = x \cap y$.

Let

$f:[0;l] \to \mathbb{R}^+ $ s.t. $f(\bar{y}) = \lvert\{y=\bar{y} \} \cap A\rvert $;

$g:[0;l] \to \mathbb{R}^+$ s.t. $g(y) =\int_{0}^{y}f(\bar{y})d\bar{y}$;

Clearly $g(0)=0$ and $g(l) = Area(A)$ and $g$ is continuous since it is an integral function. By the intermediate value theorem there exists $a \in [0;l]$ s.t. $g(a)= Area(A)/2$. And then the line $r: y=a$ has the required property.

To prove that it is unique: by contradiction suppose there exist $a_1, a_2 \in [0;l]$ s.t. $a_2 > a_1$ and $g(a_2) = g(a_1) = Area(A)/2$. Then $$0 = g(a_2) -g(a_1) = \int_{a_1}^{a_2} f(y)dy \Rightarrow f(y) \overset{\text{q.o}}{=} 0 \text{ in } [a_1; a_2]$$ Then there exists a point $\bar{a} \in [a_1; a_2]$ s.t. $f(\bar{a})=|\{y=\bar{a} \} \cap A|=0$.

Then set $B_{\bar{a}} =(\{y=\bar{a} \} \cap A)$ has measure $0$, if it is the $\emptyset$ then $A$ is not connected. If it is not then it can't be open. Then there is a point $x_0 \in B_{\bar{a}}$ s.t. it doesn't exist $\varepsilon >0$ s.t. $(x_0-\varepsilon, x_0+\varepsilon) \subset B_{\bar{a}}$ and then it doesn't exists a $B_{\varepsilon}((x_0,\bar{a})) \subset A$ and then also $A$ is not open.

For the part (ii) how can I prove continuity of $g$ (defined in the other post)?

N.Bach
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Bremen000
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  • Your proof is too complicated. You should put $g(y) = \int_{\mathbb{R}^2} 1_{[0,1]\times[0,y]\cap A} dm$, where $m$ is the Lebesgue measure and I assume $A \subset [0,1]^2$. You can see that $g$ is continuous by showing that $|g(x)-g(y)| \le |x-y|$. The choice of direction is unimportant, you can pick any one WLOG. – Olivier Jun 01 '17 at 11:48
  • @Olivier Ok I see, the complication is all about the direction, it is very faster using the indicator function. What about the part of the uniqueness? The $g$ I was talking about is the $g$ defined in the original post used to prove part (ii). I can't see how to prove it is continuous... – Bremen000 Jun 01 '17 at 12:14
  • Use the fact that $A$ is open to show that $g$ is strictly increasing. – Olivier Jun 01 '17 at 12:32

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