Bisector of two points in a Riemannian manifold has measure $0$ Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?
I was thinking about this problem with a colleague and we believe we can prove it when $M$ is a closed manifold, but even in that case the proof was much more complicated than expected. I encountered this problem while thinking about this question.
 A: Here is my attempt: it is incomplete, because it's missing the proof that the function defined in \eqref{1} below indeed fails to be differentiable on $\partial \mathcal{B}(p,q)$. Perhaps it's useful regardless of this gap.
The function
\begin{equation}
\tag{1}\label{1}
x \in M \mapsto \lvert d(x,p) - d(x,q) \rvert
\end{equation}
is Lipschitz, with zero set $\mathcal{B}(p,q)$. I think this is not differentiable on $\partial \mathcal{B}(p,q)$—except perhaps where one of $x \mapsto d(x,p)$ or $d(x,q)$ are not differentiable.
By Rademacher's theorem, one would therefore have
\begin{equation}
\mathcal{H}^n(\partial \mathcal{B}(p,q)) = 0.
\end{equation}
Next we prove that $\mathcal{B}(p,q)$ has empty interior, whence $\mathcal{B}(p,q) = \partial \mathcal{B}(p,q)$ and the proof would be concluded.
Suppose otherwise, and let $U \subset \mathcal{B}(p,q)$ be an open set. As the cut locus of both $p$ and $q$ has Hausdorff dimension $n-1$, there is a point $z \in U$ that does not belong to either. There are two minimizing geodesics $\gamma_p$ and $\gamma_q$ connecting $z$ to $p$ and $q$ respectively. As $z$ is not in the cut locus of either $p$ or $q$, we can continue both geodesics a small amount beyond $z$, while retaining their minimizing properties. Denote these extensions still by $\gamma_p$ and $\gamma_q$, parametrized so that $\gamma_p(0) = p$, $\gamma_q(0) = q$.
Assume in what follows that $\gamma_p$ and $\gamma_q$ intersect transversely at $z$—otherwise the argument is similar, but a bit easier.
Let $D = d(z,p) = d(z,q)$. Then $\gamma_p(D + \epsilon) \in U$, and by assumption
\begin{equation}
D + \epsilon = d(\gamma_p(D + \epsilon),p) = d(\gamma_p(D + \epsilon),q).
\end{equation}
The curve obtained by concatenating $\gamma_q$ from time $0$ to $D$ with $\gamma_p$ from time $D$ to $D+\epsilon$ is a path from $q$ to $\gamma_p(D+\epsilon)$ with length $D+\epsilon$. As $d(\gamma_p(D+\epsilon),q) = D+\epsilon$, it would therefore be a minimizing geodesic. However, it is not even smooth!
A: Here is a short proof supposing that $M$ is complete (if not the statement is false, see the last paragraph) using the idea from Leo Moos' answer of using Rademacher's theorem.
Suppose $\mathcal{B}(p,q)$ has measure $>0$. By Lebesgue's density theorem, almost all points $a\in\mathcal{B}(p,q)$ satisfy that $\mathcal{B}(p,q)$ has density $1$ at $a$. Fix such a point $a$, and by Rademacher's theorem suppose that $d_p:y\mapsto d(p,y)$ and $d_q:y\mapsto d(q,y)$ are smooth at $a$. Let $\gamma_p,\gamma_q$ be minimizing geodesics from $p$ and $q$ to $a$, so that $\gamma_p(0)=p,\gamma_q(0)=q$ and letting $k=d(p,a)=d(q,a)$, we have that $\gamma_p(k)=\gamma_q(k)=a$.
Let $v_p, v_q$ be the gradients of $d_p,d_q$ at $a$. Then $\lvert v_p\rvert,\lvert v_q\rvert\leq1$ due to $d_p,d_q$ being $1$-Lipschitz. Also note that $d_p(\gamma_p(t))=t$ and $d_q(\gamma_q(t))=t$ for all $t\in[0,k]$, so $\langle v_p,\gamma_p'(k)\rangle=\langle v_q,\gamma_q'(k)\rangle=1$, so $v_p=\gamma_p'(k),v_q=\gamma_q'(k)$. Moreover we cannot have $\gamma_p'(k)=\gamma_q'(k)$, because then by uniqueness of geodesics we would have $p=q$. So $v_p\neq v_q$.
Let $S_aM=\{v\in T_aM;|v|=1\}$. Then for any vector $v\in S_aM$ with $\langle v_p,v\rangle\neq\langle v_q,v\rangle$ we have $\left.\frac{d}{dt}\right|_{t=0}d(p,exp_a(tv))=\langle v_p,v\rangle\neq\langle v_q,v\rangle=\left.\frac{d}{dt}\right|_{t=0}d(q,exp_a(tv))$.
So if we define $\varepsilon_v:=\inf\{t>0;\text{exp}_a(tv)\in\mathcal{B}(p,q)\}$, then $\varepsilon_v>0$ for almost all $v\in S_aM$, and the function $v\mapsto\varepsilon_v$ is measurable (because $\{v\in T_aM;\text{exp}_a(v)\in\mathcal{B}(p,q)\}$ is closed), so there is some $\varepsilon>0$ and some set $X$ of positive measure in $S_aM$ such that exp$(tv)\not\in\mathcal{B}(p,q)$ $\forall v\in X,\forall t\in(0,\varepsilon)$. This contradicts the fact that $\mathcal{B}(p,q)$ has density $1$ at $a$.
If $M$ is not complete, the statement is false: consider the manifold $\mathbb{R}^2\setminus\{(x,0);x\geq0\}$ with the usual metric of $\mathbb{R}^2$ and let $p=(1,-2), q=(2,-1)$. Then all points $(x,y)$ with $x,y>0$ are at the same distance of $p$ and $q$.
