Ham sandwich theorem. Recently, I have been reading the Borsuk-Ulam theorem from Hatcher's algebraic topology book. In this book, there is an exercise as follows.
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.
I have some doubts regarding the above statement; each $A_i$ divided equal measure this part I did not understand properly. $\mathbb{R}^3$ has its won Lebesgue measure say $\mu$ if I take one of $A_i$ suppose $A_1$ say a unit circle $S^1$ which is compact and $\mu(S^1)=0$ then no need to divided by two equal measure. Is this statement say when I am taking one of them as the lesser dimension, we must take the Lebesgue measure concerning that dimension?
Can anyone suggest books or good references where I can find the statement of the ham sandwich theorem and its proofs?

He does not say anything about the uniqueness of the plane. It can be the case that there is an infinite amount of planes such as $A_1=A_2=A_3=B_1(0)$, where $B_1(0)$ denotes the unit ball in $\mathbb{R}^3$. Each plane $P$ with $0\in P$ will suffice in this example. You outlined another example where an infinite number of possible planes divide the sets to equal measure. This happens in your example because the sets have a zero measure.
Hint for the exercise(Edit): Let $(\mathbb{R}^3)^*$ be the dual vector space of $\mathbb{R}^3$ and investigate the properties of the maps $f_i:(\mathbb{R}^3)^* \to \mathbb{R}$, given by $\omega \mapsto \mu(\{x\in A_i: \omega(x)>r_\omega\})-\mu(\{x\in A_i: \omega(x)\le r_\omega\}).$ Choose $r_\omega\in \mathbb{R}$ in a continues way so $\mu(\{x\in A_1: \omega(x)>r\})-\mu(\{x\in A_1: \omega(x)\le r\})=0$ for all $\omega$.
