A subset $S$ of the unit sphere $\mathbb{S}^2$ is called orthogonal-pair-free if and only if there do not exist two distinct points $u, v \in S$ at distance $\frac{\pi}{2}$ from each other. Witsenhausen \cite{witsenhausen} asked the following question: {\it What is the least upper bound $\alpha_3$ on the Lesbegue measure of any measurable orthogonal-pair-free subset of $\mathbb{S}^2$?} We prove the following result in this paper: Let $\mathcal{A}$ be the collection of all orthogonal-pair-free sets $S$ such that $S$ consists of a finite number of mutually disjoint convex sets. Then, $\alpha_3 = \limsup_{S \in \mathcal{A}} \mu(S)$. Thus, if the double cap conjecture \cite{kalai1} is not true, there is a set in $\mathcal{A}$ with measure strictly greater than the measure of the double cap.

翻译：单位球面$\mathbb{S}^2$的子集$S$称为无正交对集合，当且仅当不存在两个不同点$u, v \in S$使得它们之间的距离为$\frac{\pi}{2}$。Witsenhausen \cite{witsenhausen}提出了以下问题：{\it 在$\mathbb{S}^2$的任何可测无正交对子集上，Lesbegue测度的最小上界$\alpha_3$是多少？}本文证明以下结果：设$\mathcal{A}$为所有由有限个互不相交凸集构成的无正交对集合$S$的全体。那么，$\alpha_3 = \limsup_{S \in \mathcal{A}} \mu(S)$。因此，如果双帽猜想\cite{kalai1}不成立，则存在$\mathcal{A}$中的集合，其测度严格大于双帽的测度。