I am currently reading the following lecture notes: http://www-math.mit.edu/~goemans/18409-2006/lec6.pdf. They prove the following:
Lemma 4: Consider a function $f: S_{n-1}\rightarrow\mathbb{R}$ which is 1-Lipschitz. We define $m(f) \in \mathbb{R}$, the median of $f$, such that $\mu(A^+)\geq\frac{1}{2}$ and $\mu(A^-)\geq\frac{1}{2}$ where $A^+ = \{x:f(x)\geq m(f)\}$ and $A^- = \{x:f(x)\leq m(f)\}$. Then,
$$\mu(\{x:|f(x)-m(f)|>\epsilon\}) \leq (1+o(1))e^{-\frac{\epsilon^2n}{2}}.$$
In reading their proof of Lemma 4, it occurred to me that we could make an equivalent argument for another pair of subsets, $B^+$ and $B^-$, where $\mu(B^+) = \mu(B^-) = \frac{1}{2}$. (i.e. replace $A^+$ with $B^+$ and $A^-$ with $B^-$ in their proof of Lemma 4.) Thus, we prove that the measure concentrates on $B^+ \cap B^-$, but we also have that the measure concentrates on $A^+ \cap A^-$. If we choose $B^+ \cap B^-$ to have no intersection with $A^+ \cap A^-$ and take the dimension of the space to infinity, we get that the measure concentrates in two distinct subsets of $S_{n-1}$, simultaneously. How is this possible?