Proving the Ham-Sandwich theorem for $n=3$ 
Let $A_1, A_2, A_3$ be compact sets in $\Bbb R^3$. Use the Borsuk–Ulam theorem to show that there is one plane $P \subset \Bbb R^3$ that simultaneously divides each $A_i$ into two pieces of equal measure.

Every point $s \in \Bbb S^2$  defines a unit vector in $\Bbb R^3$ which can be used to orient a plane $P \subset \Bbb R^3$. So let $s \in \Bbb S^2$ be such that the plane $P_s \subset \Bbb R^3$ it splits $A_1$ in two parts of equal volume.
If $f_1(s)= \text{measure of } A_2 \text{ in front of } P_s - \text{measure of } A_2 \text{ in behind of } P_s$ and $f_2(s) = \text{measure of } A_3 \text{ in front of } P_s - \text{measure of } A_3 \text{ in behind of } P_s$.
Now if $f :\Bbb S^2 \to \Bbb R^2$ is defined by $f(s)=(f_1(s), f_2(s))$, then by the Borsuk-Ulam theorem there exists $s' \in \Bbb S^3$ for which we have that $f(s')=f(-s')$ and so $P_{s'}$ is the plane that we're looking for.
How can I make the definitions for $f_1$ and $f_2$ precise? I think I ought to use the Lebesgue measure for those?
 A: Let's first make sure we know how we're defining $P_s$. For each $s\in S^2$ we get a line, say $\ell_s$ passing through $s$ and the origin. For each point, $q$ on $\ell_s$, we get a plane, say $P_s^q$ that contains $q$ and is orthogonal to $\ell_s$. Each plane $P_s^q$ splits $\mathbb{R}^3$ into two components, one which is on the side of $P_s^q$ that is in the direction that the ray from the origin to $s$ points; we think of this as the component in front of $P_s^q$. Let's give the name $F_s^q$ to this component of $\mathbb{R}^3\backslash P_s^q$. Sliding $q$ along $\ell_s$ we have a continuous function of $q$ given by $m(A_1\cap F_s^q)$, where $m$ is the Lebesgue measure. Then, since $A_1$ is compact, so everything is finite, the intermediate value theorem tells us there is some $q$ such that $m(A_1\cap F_s^q)=m(A_1)/2.$ Then, we define $P_s:=P_s^q$ where we choose $q$ to be satisfying that equality. (If the sets are suitably nice, there will be a unique such $q$, though there's slightly more to say otherwise).
Now that we've got a well defined $P_s$, let's define $F_s$ to be the component of $\mathbb{R}^3\backslash P_s$ that is on the side of $P_s$ in the direction that $s$ points from the origin (as above) and $B_s$ to be the other component of $\mathbb{R}^3\backslash P_s$. Then, we define $f_1(s)=m(A_2\cap F_s)-m(A_2\cap B_s)$ and $f_2(s)=m(A_3\cap F_s)-m(A_3\cap B_s)$.
