Probability measure on partition theorem

Question

Probability measure in $\mathbb{R}^1$: Continuous function $f$ that is positive everywhere $\ \int_{-\infty}^{\infty} f dx =1$ (total area under the curve equals 1). For visualization see here. Measure of a subset $S\subseteq \mathbb{R}^1$: Integral over that subset. There is a way to split measure into two equal parts(1/2 and 1/2)

Probability measure in $\mathbb{R}^2$: Continuous function $f:\mathbb{R}^2\to \mathbb{R}$ that is positive everywhere and $\ \iint_\mathbb{R^2}f d\textbf{x}d\textbf{y}=1.$ For visualization see here.

Equipartition theorem (Ceder 1964): Every probability measure $f$ in $\mathbb{R}^2$ can be divided into 6 equal parts by 3 concurrent lines.

Proof: For every direction $\vec{v}$ there exists a unique line $\ell$ parallel to $\vec{v}$ that partitions $f$ into two equal parts. $\ell$ varies continuously with $\vec{v}.$ Given $\vec{v}$ and given a point $x\in \ell(\vec{v})$ there exist 4 unique rays from $\ell$, two on each side, that partition $f$ into 6 equal parts. As $x$ moves forwards, the rays rotate backwards. For visualization see here. There is a unique point $x^*$ for which $r_1,r_4$ form a straight line. Let $\alpha$ be the clockwise angle from $r_2$ to $r_3$.$\alpha =\alpha(\vec{v})$ varies continuously with $\vec{v}.$ When $\vec{v}$ turns $180°$, $x^∗$ returns to the same place, $$r_1↔r_4, r_2↔r_3, α↔360°-α.$$ There must exist an intermediate value of $\vec{v}$ for which $\alpha(\vec{v})=180°.$ For visualization see here. Q.E.D

My question is how can Prove that every probability measure in $\mathbb{R^2}$ can be divided into 6 equal parts by 3 lines, 2 of which are parallel? For visualization see here.We can use the above Equipartition theorem.

Answer 1

You can argue using the intermediate value theorm.

Here is a sketch:

Claim 1. For each $\theta\in(-\pi/2,\pi/2)$, there is one and only one way to divide the plane into three equal-measure regions using two parallel lines $\ell_1(\theta)$ (the bottom one) and $\ell_2(\theta)$ (the top one) with angle $\theta$.

We denote the obtained three regions by $Q_T$ (top), $Q_M$ (middle), and $Q_B$ (bottom).

Claim 2. For every $\theta\in(-\pi/2,\pi/2)$, there is one and only one line $\ell_3(\theta)$ that divides each of the two regions $Q_T$ and $Q_B$ into equal-measure parts.

The line $\ell_3$ divides $Q_M$ into two parts, which we denote by $Q_{ML}$ (the left one) and $Q_{MR}$ (the right one).

Claim 3. The triplet $\big(\ell_1(\theta),\ell_2(\theta),\ell_3(\theta)\big)$ is continuous in $\theta$.

Claim 4. We can extend $\ell_1,\ell_2,\ell_3$ to the closed interval $[-\pi/2,\pi/2]$ in a continuous fashion. At $\theta=-\pi/2$ and $\theta=\pi/2$, we get the same three lines and the same six regions, but $\ell_1$ and $\ell_2$ switch places and so do $Q_{ML}$ and $Q_{MR}$.

Namely, at $\theta=-\pi/2$:

• $\ell_1$ is the left line and $\ell_2$ the right one,
• $Q_{ML}$ is the top middle region and $Q_{MR}$ is the bottom middle region,

while at $\theta=\pi/2$, it is the other way around.

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Now, let \begin{align*} f(\theta):=\mu\big(Q_{MR}(\theta)\big) - \mu\big(Q_{ML}(\theta)\big) \end{align*} where $\mu(A)$ denotes the measure of a region $A$. Note that $f(\theta)$ is continuous on $[-\pi/2,\pi/2]$ with $f(-\pi/2)=-f(\pi/2)$ (because $Q_{ML}$ and $Q_{MR}$ switch places). Therefore, by the intermediate value theorem, there must exist a point $\theta_0\in[-\pi/2,\pi/2]$ at which $f(\theta_0)=0$.

At $\theta=\theta_0$, the three lines $\ell_1,\ell_2,\ell_3$ divide the plane into six equal-mass regions.