Let $(\Omega,d)$ be a compact metric space and $\mathcal P(\Omega)$ its space of Borel probability measures. Let $D=\{ d_p\mid p\in\Omega\}$ where $d_p(x)=d(p,x)$ be the set of all "distance functionals". As usual, we can think of $D$ acting on $\mathcal P(\Omega)$ (or vice versa) via integration i.e. $\langle d_p,\mu\rangle = \int_\Omega d_p(x)\,\mathrm d\mu(x)$.

Title Question

Does $D$ acting on $\mathcal P(\Omega)$ via integration separate points?

Or equivalently,

If $\mu,\nu \in \mathcal P(\Omega)$ and $\langle d_p,\mu\rangle = \langle d_p,\nu\rangle$ for all $p\in \Omega$, then must $\mu=\nu$?

Alternative Formulations

There are a few other ways to frame the question as well.

Probabilistic Formulation

Rewriting all integrals as expectations the question becomes,

If $\mathbb E_{X\sim\mu}[d_p(X)] = \mathbb E_{Y\sim\nu}[d_p(Y)]$ for all $p\in \Omega$, then must $\mu=\nu$?

In other words, does knowing the expected distance to a point for all points determine the measure?

Geometric Formulation

Recall that the 1-Wasserstein distance is given by $W_1(\mu,\nu) = \inf_{\gamma\in\Gamma(\mu,\nu)} \int_{\Omega\times\Omega} d(x,y) \,\mathrm d\gamma(x,y)$ where $\Gamma(\mu,\nu)$ is the set of couplings between $\mu$ and $\nu$ i.e. Borel probability measures on $\Omega\times\Omega$ with marginals $\mu$ and $\nu$ respectively. Since the product measure $\delta_p\otimes\mu$ is the unique coupling between a Dirac delta measure $\delta_p$ and $\mu$, we have that

$$W_1(\delta_p,\mu)=\int_{\Omega\times\Omega} d(x,y)\,\mathrm d(\delta_p\otimes\mu)(x,y)=\int_\Omega d(p,y)\,\mathrm d\mu(y)=\langle d_p,\mu\rangle$$

Now the question can be stated geometrically as

If $W_1(\delta_p,\mu)=W_1(\delta_p,\nu)$ for all $p\in \Omega$, then must $\mu=\nu$?

In other words, does knowing the $W_1$ distance to the extreme points of $\mathcal P(\Omega)$ completely determine the probability measure?

Integral Transform Formulation

Define the distance transform of $\mu\in\mathcal P(\Omega)$ as the function $\phi_\mu:\Omega\to\mathbb R$ given by $\phi_\mu(p) = \int_\Omega d(p,x)\,\mathrm d\mu(x)$. The question can now be restated as,

Is the distance transform injective on $\mathcal P(\Omega)$?

Moreover, by the geometric formulation we have $\phi_\mu(p) = W_1(\delta_p,\mu)$. We will use the weak-$*$ topology for $\mathcal P(\Omega)$ (which coincides with the $W_1$ topology). Since the map $p\mapsto \delta_p$ is an embedding of $\Omega$ into $\mathcal P(\Omega)$, it follows that $\phi_\mu:\Omega\to\mathbb R$ is continuous. Denote the distance transform by $\Phi(\mu)=\phi_\mu$. Since $\mathcal P(\Omega)$ is compact Hausdorff and $C(\Omega)$ is Hausdorff we can restate the question as

If $\Phi:\mathcal P(\Omega)\to C(\Omega)$ is continuous, is it an embedding?

Final Thoughts

Are any of these equivalent statements true? I have unfortunately only been able to reformulate the question and have not identified any clear proof, though I wouldn't be surprised if there is an easy one I'm overlooking. The geometric formulation of the problem leads me to believe that $D$ does indeed separate points in $\mathcal P(\Omega)$. However, if the answer is affirmative then I feel the resulting nice properties of $\Phi$ would make it something that would be easy to look up. Any insight would be appreciated.

Update: In light of George Lowther's elegant 4-point counter-example and Pietro Majer's affirmative answer for $\Omega=[0,1]$, it would be interesting to better understand what factors determine whether the underlying metric space yields an affirmative answer.

George's counter-example can be extended to counter-examples where $\Omega$ is a sphere (with intrinsic metric). Thus, requiring $\Omega$ to be positive-dimensional, a manifold, connected, path-connected, simply-connected, etc, will not make the issue go away. On the other hand, Pietro suspects that the answer is again affirmative in the case when $\Omega$ is a compact convex subset of Euclidean space.

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    $\begingroup$ I don't know the answer, but one can reduce the problem to the case that $\Omega$ is finite, since probability measures with finite support are weak*-dense. $\endgroup$ Commented Aug 17, 2020 at 8:24
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    $\begingroup$ Isn't this a consequence of the Riesz representation theorem and the Stone-Weierstraß approximation theorem? $\endgroup$ Commented Aug 17, 2020 at 8:45
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    $\begingroup$ But Stone-Weierstraß only implies that the algebra generated by the distance functionals is dense. One needs the weak$^*$ density of the generated subspace. $\endgroup$ Commented Aug 17, 2020 at 9:15
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    $\begingroup$ @JochenWengenroth: You're right, of course. I confused a few arguments and thus thought this would not be a problem; but of course, it actually is a problem... $\endgroup$ Commented Aug 17, 2020 at 9:29
  • $\begingroup$ See also this answer: math.stackexchange.com/a/90511/442 $\endgroup$ Commented Mar 26, 2023 at 10:42

2 Answers 2


No. Suppose that $\Omega$ consists of four points arranged in a square, where adjacent points have distance 1 between them and opposite points have distance 2. Specifically, if the points are labeled A,B,C,D then \begin{align} & d(A,C)=d(B,D)=2,\\ & d(A,B)=d(B,C)=d(C,D)=d(D,A)=1. \end{align} For example, A,B,C,D could be equally spaced around a circle, using the internal circle metric.

There are precisely two probability measures assigning probability 1/2 to each of two opposite points and probability zero to the remaining two points. \begin{align} & \mu(\{A\})=\mu(\{C\}) = 1/2,\ \mu(\{B,D\})=0,\\ & \nu(\{B\})=\nu(\{D\})=1/2,\ \nu(\{A,C\})=0. \end{align} You can check that these two measures give the same integral for all `distance functions'. The average distance from every point is equal to 1 under both of these.

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    $\begingroup$ Elegant counter-example.This idea can be pushed a bit further as well though it was perhaps implicit: The empirical measure of a pair of antipodal points on the unit circle $\delta_p / 2+\delta_{-p} / 2$ produces an expected intrinsic distance of $\pi/2$ to any point on the circle. Thus, any two "empirical measure of antipodal points" cannot be separated by distance functionals. This shows the obstruction isn't simply due to finiteness of $\Omega$ and that the obstruction can't be avoided by requiring the underlying space to be nonzero dimensional, a manifold, connected, path-connected, etc. $\endgroup$ Commented Aug 18, 2020 at 8:29
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    $\begingroup$ Actually, this idea using antipodal points works for spheres as well. Thus being simply-connected is not a sufficient condition to ensure an affirmative answer either. $\endgroup$ Commented Aug 18, 2020 at 8:48

On the positive side, the answer is affirmative if $\Omega$ is the unit interval $[0,1]$ with its standard distance. In this case $\phi_\mu$ is a convex $1$-Lipschitz function (in fact, it is also defined for all $p\in\mathbb{R}$, with $\phi'(p)=\mathrm{sgn} p$ for $p\notin[0,1]$), with left and right derivatives $$\phi_-'(p)=\mu[0,p)-\mu[p,1]= 2\mu[0,p)-1$$ $$\phi_+'(p)=\mu[0,p] -\mu (p,1]= 2\mu[0,p] -1=1-2\mu(p,1]$$ so that $\mu$ is determined on all intervals, hence on all Borel subsets.

Conversely, note that any convex function $\phi$ as above
may be written in the form $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ for some Borel probability measure $m$ on $[0,1]$. This because $g:= \frac{1}{2}\big(1-\phi_+'\big) $ is a nonnegative bounded cadlag function, so there is a Borel probability function $m$ such that $g(p)=m(p,1]$, whence $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ follows easily from the above relations.

I'd guess the answer is also affirmative for $\Omega$ a convex compact set of $\mathbb{R}^n$ with the Euclidean distance.


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