I am try to solve the exercise 1.1.9 of Hatcher book about the ham sandwich theorem that says:
Let $A_1, A_2, A_3$ be compcat 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'm trying to solve this exercise based on some solutions I read on the internet and I'm having trouble justifying some statements.
FIRST STEP
Let $x \in \mathbb{S}^2$ and $B_t = \{ y \in \mathbb{R}^3 : x\cdot y \geq t \}$, where $t\in \mathbb{R}$, a subset of $\mathbb{R}^3$ consisting of a region of space on one side of the plane through the line $t x$ with normal vector $x$.
Define $g_t : \mathbb{S}^2 \times \mathbb{R} \to \mathbb{R}^3$ by
$$ g_t(x) = \left( m(A_1 \cap B_t) - m(A_1 \cap B_t^{c}), m(A_2 \cap B_t) - m(A_2 \cap B_t^{c}),m(A_3 \cap B_t) - m(A_3 \cap B_t^{c}) \right), $$
where $m$ is the Lebesgue mensure. I'm trying to show that $g_t$ is a continuous function, but I don't know how I'm going to do that, since the function depends on the concept of measure and I don't know how to define continuity of measures of sets.
Any idea? This function is in fact continuous? How I can prove this?
Thanks in advance.
Edited and rewritten. I realise that $g : \mathbb{S}^2 \times \mathbb{R} \to \mathbb{R}^3$ is actually a function of two variables, so my previous answer wasn't appropriate.
To distil this problem to its essentials, I suggest we instead study the function $f : \mathbb{S}^2 \times \mathbb{R} \to \mathbb R$ given by $f(\mathbf x, t) = m(E_{\mathbf x, t})$, where $E_{\mathbf x, t}$ is the set $$ E_{\mathbf x, t} = A \cap \{\mathbf y \in \mathbb R^3 : \mathbf x . \mathbf y \geq t \},$$ and $A$ is a compact subset of $\mathbb R^3$. If we can prove that $f$ is continuous, then certainly we've got the techniques to prove that $g$ is continuous too.
Probably the easiest approach is to consider a sequence $(\mathbf x_n, t_n)_{n \in \mathbf N}$ such that $\lim_{n \to \infty} (\mathbf x_n, t_n) = (\mathbf x_\star, t_\star)$. Our task is then to prove that $\lim_{n \to \infty} f(\mathbf x_n, t_n) = f(\mathbf x_\star, t_\star)$.
To compute this limit, we're going to use the Dominated Convergence Theorem. For any measurable subset $E \subset \mathbb R^n$, let $\chi_E$ be the indicator function on this subset.
The sequence of functions $\chi_{E_{\mathbf x_n, t_n}}$ converges pointwise to the function $\chi_{E_{\mathbf x_\star, t_\star}}$. [Technically, this is not necessarily true on the hyperplane $\{\mathbf y \in \mathbb R^3 : \mathbf x_\star . \mathbf y = t_\star \} $; however, this doesn't bother us because this hyperplane is a set of measure zero.]
The function $\chi_{A}$ dominates the sequence. $\chi_A$ is integrable. To see this, we observe that $A$ is bounded by Heine-Borel, and therefore, $m(A) < \infty$.
By Dominated Convergence, we have $$ \lim_{n \to \infty} f(\mathbf x_n, t_n) = \lim_{n \to \infty} \int \chi_{E_{\mathbf x_n, t_n}} dm = \int \chi_{E_{\mathbf x_\star, t_\star}} dm = f(\mathbf x_\star, t_\star),$$ as required.