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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 $v_p\neq v_q$.

Now note that for any vector $v\in T_aM$ 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 for almost all $v\in T_aM$ there is some $\varepsilon_v>0$ such that $exp_a(tv)\not\in\mathcal{B}(p,q)$, which easily 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 $q$.

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 $q$.

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 $|v_p|,|v_q|\leq1$ 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 $v_p\neq v_q$.

Now note that for any vector $v\in T_aM$ 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 for almost all $v\in T_aM$ there is some $\varepsilon_v>0$ such that $exp_a(tv)\not\in\mathcal{B}(p,q)$, which easily 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 $q$.

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 $v_p\neq v_q$.

Now note that for any vector $v\in T_aM$ 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 for almost all $v\in T_aM$ there is some $\varepsilon_v>0$ such that $exp_a(tv)\not\in\mathcal{B}(p,q)$, which easily 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 $q$.

Here is a short proof in the case $M$ 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 $|v_p|,|v_q|\leq1$ due to $d_p,d_q$ being $1$-Lipschitz. Combining this with the fact that $d_p(\gamma_p(t))=t$ and $d_q(\gamma_q(t))=t$ for all $t\in[0,k]$, we must have $v_p=\gamma_p'(k),v_q=\gamma_q'(k)$. Also we cannot have $\gamma_p'(k)=\gamma_q'(k)$, because then by uniqueness of geodesics we would have $p=q$. So $v_p\neq v_q$.

Now note that for any vector $v\in T_aM$ 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 for almost all $v\in T_aM$ there is some $\varepsilon_v>0$ such that $exp_a(tv)\not\in\mathcal{B}(p,q)$, which easily 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 $\mathbb{R}^2\setminus\{(x,0);x>0\}$ with the usual metric of $\mathbb{R}^2$ and let $p=(3,-4),q=(0,-5)$. Then all points $(x,y)$ with $x,y>0$ are at the same distance of $p$ and $q$.

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 $|v_p|,|v_q|\leq1$ 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 $v_p\neq v_q$.

Now note that for any vector $v\in T_aM$ 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 for almost all $v\in T_aM$ there is some $\varepsilon_v>0$ such that $exp_a(tv)\not\in\mathcal{B}(p,q)$, which easily 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 $q$.