Prove the Ham Sandwich theorem.
Solution
Goal: Let \(A_1, A_2, \dots, A_n \subset \mathbb{R}^n\) be \(n\) bounded Lebesgue measurable subsets of finite measure \(\mu(A_i) < \infty\). Prove the Ham Sandwich Theorem: there exists an affine hyperplane \(H \subset \mathbb{R}^n\) that simultaneously bisects all \(n\) sets, i.e., divides each \(A_i\) into two pieces of equal measure.
1. Parameterize oriented hyperplanes via the sphere \(S^n\).
1.1. Embed \(\mathbb{R}^n\) into \(\mathbb{R}^{n+1}\) at height 1: \(\iota(x) = (x, 1) \in \mathbb{R}^{n+1}\).
1.2. For each unit vector \(u = (v, w) \in S^n \subset \mathbb{R}^{n+1}\) (with \(v \in \mathbb{R}^n, w \in \mathbb{R}\)), define the closed affine half-space in \(\mathbb{R}^n\): \[H^+(u) = \{ x \in \mathbb{R}^n \mid \langle \iota(x), u \rangle \ge 0 \} = \{ x \in \mathbb{R}^n \mid \langle x, v \rangle + w \ge 0 \}.\]
1.3. The opposite vector \(-u\) gives the complementary half-space: \[H^+(-u) = \{ x \in \mathbb{R}^n \mid \langle x, -v \rangle - w \ge 0 \} = \{ x \in \mathbb{R}^n \mid \langle x, v \rangle + w \le 0 \}.\]
1.4. The boundary hyperplane is \(H(u) = \{ x \in \mathbb{R}^n \mid \langle x, v \rangle + w = 0 \}\).
1.5. Since \(H(u)\) has Lebesgue measure zero in \(\mathbb{R}^n\), \(\mu(A_i \cap H(u)) = 0\), so: \[\mu(A_i \cap H^+(u)) + \mu(A_i \cap H^+(-u)) = \mu(A_i) \quad \text{for each } i \in \{1, \dots, n\}.\]
1.6. The two half-spaces \(H^+(u)\) and \(H^+(-u)\) are complementary, overlapping only on the hyperplane \(H(u)\); since \(H(u)\) has measure zero, the measures of the two intersections add to \(\mu(A_i)\).
2. Define a continuous map \(F \colon S^n \to \mathbb{R}^n\).
2.1. For each \(i \in \{1, \dots, n\}\), define the \(i\)-th coordinate function \(f_i \colon S^n \to \mathbb{R}\) by: \[f_i(u) = \mu(A_i \cap H^+(u)).\]
2.2. \(f_i\) is continuous: Let \(u_k \to u\) in \(S^n\).The characteristic functions \(\chi_{A_i \cap H^+(u_k)}\) converge almost everywhere to \(\chi_{A_i \cap H^+(u)}\) (except possibly on the hyperplane \(H(u)\), which is a null set).Since \(|\chi_{A_i \cap H^+(u_k)}| \le \chi_{A_i}\) and \(\mu(A_i) < \infty\), the Dominated Convergence Theorem implies: \[\lim_{k \to \infty} f_i(u_k) = \lim_{k \to \infty} \int_{\mathbb{R}^n} \chi_{A_i \cap H^+(u_k)} \, d\mu = \int_{\mathbb{R}^n} \chi_{A_i \cap H^+(u)} \, d\mu = f_i(u).\]
2.3. Define \(F \colon S^n \to \mathbb{R}^n\) by \(F(u) = (f_1(u), f_2(u), \dots, f_n(u))\).
2.4. Since each component \(f_i\) is continuous, \(F\) is continuous.
2.5. The Dominated Convergence Theorem applies because the integrands are dominated by \(\chi_{A_i}\) and \(\mu(A_i) < \infty\), so each \(f_i\) is continuous; hence \(F\) is continuous.
3. Apply the Borsuk-Ulam theorem to \(F\).
3.1. By the Borsuk-Ulam theorem (proved in P-AMD-6OJQMSOZ for \(n=2\), and holding for all \(n \ge 1\)), for any continuous map \(F \colon S^n \to \mathbb{R}^n\), there exists a point \(u^* \in S^n\) such that: \[F(u^*) = F(-u^*).\]
3.2. In terms of coordinate functions, this means for all \(i \in \{1, \dots, n\}\): \[f_i(u^*) = f_i(-u^*) \iff \mu(A_i \cap H^+(u^*)) = \mu(A_i \cap H^+(-u^*)).\]
3.3. Combining with the total measure equality from 1: \[\mu(A_i \cap H^+(u^*)) + \mu(A_i \cap H^+(-u^*)) = \mu(A_i) \implies 2 \mu(A_i \cap H^+(u^*)) = \mu(A_i) \implies \mu(A_i \cap H^+(u^*)) = \frac{1}{2} \mu(A_i).\]
3.4. If \(v^* = 0\), then \(u^* = (0, \pm 1)\), giving \(H^+(u^*) = \mathbb{R}^n\) or \(\emptyset\), which would mean \(\mu(A_i) = 0\) for all \(i\).If any \(\mu(A_i) > 0\), then \(v^* \neq 0\), so \(H(u^*)\) is a genuine affine hyperplane in \(\mathbb{R}^n\).
3.5. The hyperplane \(H(u^*)\) bisects all \(n\) sets \(A_1, \dots, A_n\) simultaneously.
3.6. Borsuk–Ulam gives \(F(u^*) = F(-u^*)\), which by <2>2–<2>3 forces \(\mu(A_i \cap H^+(u^*)) = \frac{1}{2}\mu(A_i)\) for every \(i\); hence \(H(u^*)\) bisects each \(A_i\).
4. Conclusion.
4.1. <1>1–<1>3 establish the Ham Sandwich theorem for any \(n\) measurable sets of finite measure in \(\mathbb{R}^n\).