Let \(A_1, A_2, A_3\) be compact sets in \(\mathbb{R}^3\). Use the Borsuk–Ulam theorem to show that there is one plane \(P \subset \mathbb{R}^3\) that simultaneously divides each \(A_i\) into two pieces of equal measure.
Solution
Let \(\mu\) denote three-dimensional Lebesgue measure. Since each \(A_i\) is compact, it is measurable and has finite measure.
1. Parameterize oriented affine planes and their two sides by points\[ q=(v,d)\in S^3\subset\mathbb R^3\times\mathbb R. \]For \(v\ne0\), the corresponding plane is\[ P_q=\{x\in\mathbb R^3:v\cdot x=d\}, \]with positive and negative sides determined by the sign of \(v\cdot x-d\).
Multiplying \((v,d)\) by \(-1\) leaves the underlying plane unchanged and exchanges its two sides. The two points of \(S^3\) with \(v=0\) are harmless compactification points and will be treated below.
2. For \(i=1,2,3\), define\[ F_i(v,d)=\int_{A_i}\operatorname{sgn}(v\cdot x-d)\,d\mu(x), \]where \(\operatorname{sgn}(t)=1\) for \(t>0\), \(-1\) for \(t<0\), and \(0\) for \(t=0\).Then\[ F=(F_1,F_2,F_3):S^3\to\mathbb R^3 \]is continuous.
Fix \(q=(v,d)\in S^3\) and let \(q_n=(v_n,d_n)\to q\). For each fixed \(x\) with \[ v\cdot x-d\ne0, \] one has \[ \operatorname{sgn}(v_n\cdot x-d_n) \longrightarrow \operatorname{sgn}(v\cdot x-d). \]
If \(v\ne0\), the exceptional set \[ \{x:v\cdot x=d\} \] is an affine plane and therefore has three-dimensional Lebesgue measure zero. If \(v=0\), then \(|d|=1\), so \(v\cdot x-d=-d\ne0\) for every \(x\) and there is no exceptional set. The integrands are uniformly bounded by \(1\). Dominated convergence therefore gives \[ F_i(q_n)\to F_i(q) \] for each \(i\), proving continuity of \(F\).
3. The map \(F\) is odd:\[ F(-q)=-F(q). \]
For every \(x\), \[ (-v)\cdot x-(-d)=-(v\cdot x-d), \] so the sign of the integrand reverses. Thus each \(F_i\) changes sign under \(q\mapsto-q\).
4. There exists \(q=(v,d)\in S^3\) such that\[ F(q)=0. \]
Apply the Borsuk–Ulam theorem to the continuous map \[ F:S^3\to\mathbb R^3. \] There is \(q\in S^3\) with \[ F(q)=F(-q). \] By oddness from 3, \[ F(q)=-F(q), \] hence \(F(q)=0\).
5. Unless all three sets have measure zero, the point \(q\) from 4 has \(v\ne0\).
If \(v=0\), then \(d=\pm1\). For \(d=1\) the integrand is identically \(-1\), so \[ F_i(0,1)=-\mu(A_i), \] and for \(d=-1\) it is identically \(1\), so \[ F_i(0,-1)=\mu(A_i). \] Thus \(F(q)=0\) at such a point only when every \(\mu(A_i)=0\). If all three measures are zero, any plane already bisects all three sets.
6. For the affine plane\[ P=P_q=\{x:v\cdot x=d\} \]with \(q\) as in 4 and \(v\ne0\), each \(A_i\) has equal measure on the two sides of \(P\).
An affine plane has three-dimensional Lebesgue measure zero, so \[ \mu(A_i\cap P)=0. \] The equation \(F_i(q)=0\) says exactly \[ \mu\bigl(A_i\cap\{v\cdot x>d\}\bigr) = \mu\bigl(A_i\cap\{v\cdot x<d\}\bigr). \] Thus \(P\) bisects the measure of \(A_i\) for all three indices simultaneously.