If $A_1,...,A_n$ are measurable subsets of $S^n$, then there is a great $S^{n-1}$ cutting each $A_i$ exactly in half.
The tools I have at my disposal are the Borsuk Ulam theorem and the Ham Sandwich theorem.
My idea was to reduce it to the Ham Sandwich theorem as follows. View $S^n$ as $D^n/S^{n-1}$. Let $A_1,...,A_n$ be measurable subsets of $D^n/S^{n-1}$ and let's assume that they don't contain the $S^{n-1}$ which is collapsed to a point. So the $A_i's$ are subsets of $D^n\subset \Bbb{R}^n$. By the Ham Sandwich theorem, we choose a hyperplane that cuts each $A_i$ exactly in half. In the quotient space $D^n/S^{n-1}$, this hyperplane is a great $S^{n-1}$ cutting each $A_i$ in half.
Can we make this rigorous? Is there a better solution?
Thanks.
You may just apply Borsuk-Ulam theorem directly.
Define a function $f$ from $S^n$ to $\mathbb{R}^n$ as follows:
If $x$ is a point on $S^n$, then there is a great $S^{n-1}$ that is orthogonal to the point $x$. The $S^{n-1}$ divides $S^n$ into two regions. Let's call them $U$ and $V$, where $U$ is the region containing $x$. If $\mu$ is the measure given, define $f(x)=(\mu(A_1\cup U), \ldots, \mu(A_n\cup U))$.
This is a continuous map from $S^n$ to $\mathbb{R}^n$. Thus, by Borsuk-Ulam theorem, there is a point $y$ such that $f(y)=f(-y)$. Then the great $S^{n-1}$ orthogonal to $y$ is the desired one.