Return to Answer

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $F_{p^*}$ has measure $0$.

iff

$$\mu\times\mu(\{(p,p^*)\in M\times M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\})=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has measure $0$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $M\setminus F_{p^*}$ has measure $0$.

iff

$$\mu\times\mu(\{(p,p^*)\in M\times M;\text{there is more than one minimizer of $d(p^*,\cdot)$ in }O_p\})=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has full measure in $M$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $F_{p^*}$ has measure $0$.

iff

$$\mu((p,p^*)\in M\times M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p)=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has measure $0$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $F_{p^*}$ has measure $0$.

iff

$$\mu\times\mu(\{(p,p^*)\in M\times M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\})=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has measure $0$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. This is equivalent by Fubini applied to the product measure space $M\times M$ to saying that for almost all points $p\in M$, the set $\{p^*\in M;p\in F_{p^*}\}$ has measure $0$.

That is, it will be enough to prove that for all $p\in M$, the set of points $p^*$ whose distance to $O_p$ is minimized at more than $1$ point has measure $0$. But $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $F_{p^*}$ has measure $0$.

iff

$$\mu((p,p^*)\in M\times M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p)=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has measure $0$. 

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.