Exercise 1.1.9

Proof. Since $A_3$ is compact, it is closed and bounded; thus, we can assume without loss of generality that $A_3$ lies in the region $z>0$ by changing coordinates. Then, embed $S^2$ into ${ℝ}^3$ such that the center of $S^2$ and the origin in ${ℝ}^3$ coincide; every unit vector $s\in S^2$ then defines a direction in ${ℝ}^3$. We first note that every pair $(s,r)\in S^2\times ℝ$ defines a unique plane $P(s,r)$ with normal vector $s$ distance $r$ from the origin. Note that negative $r$ denotes that the plane lies in the negative direction with respect to the direction defined by $s$. Thus, we can define $\rho (s,r)=\mu (A_3^{\text{+}}(s,r))$, the measure of the subset of $A_3$ on the positive side of the plane $P(s,r)$. $\rho (s,r)$ is continuous in $r$ since $\mu$ is continuous from above and below.

Now fix $s$. Since $\rho (s,r)=\mu (A_3)$ for $r$ sufficiently small and $\rho (s,r)=0$ for $r$ sufficiently large, by the intermediate value theorem there exists for this fixed $s$ some $r_0$ such that $\rho (s,r_0)=\frac{1}{2}\mu (A_3)$. If there are multiple such $r_0$ that form an interval $I$, then we take $r_0$ to be the midpoint of the interval $I$. Thus, for each $s$ we can define a unique plane $P(s)$ such that $\mu (A_3^{\text{+}}(s,r))=\frac{1}{2}\mu (A_3)$.

We now let $f(s)=(\mu (A_1^{\text{+}}(s)),\mu (A_2^{\text{+}}(s)))$, where $A_i^{\text{+}}(s)$ denotes the subset of $A_i$ on the positive side of $P(s)$, is continuous. $f$ is continuous since small perturbation in $s$ will only lead to small changes in our plane $P(s)$, and therefore in $\mu (A_1^{\text{+}}(s))$ and $\mu (A_2^{\text{+}}(s))$. Note that since $P(s)=-P(-s)$, i.e., the plane defined for $-s$ is the same as that defined for $s$, except oriented in the other direction, we have the equality $f(s)+f(-s)=(\mu (A_1),\mu (A_2))$.

By the Borsuk-Ulam theorem, there now exists $s_0\in S^2$ such that $f(s_0)=f(-s_0)$. But by the equality $f(s)+f(-s)=(\mu (A_1),\mu (A_2))$, we see that then $f(s_0)=f(-s_0)=\frac{1}{2}(\mu (A_1),\mu (A_2))$, and so our plane $P(s_0)$ divides the sets $A_1,A_2,A_3$ into halves of equal measure. □