One can construct a finite measure on a compact metric space $(X,d)$ by the following procedure:

Fix a non-negative sequence $\{\epsilon_n\}$, $\epsilon_n \to 0$. Let $Y_{\epsilon_n}$ be the minimal covering net: $\bigcup_{y \in Y_{\epsilon_n}} \mathcal{B} (y, \epsilon_n ) = X$ and there is no lower-cardinality net with this property. Let $\#Y_{\epsilon_n}$ denote the respective minimum cardinality.

Construct the measure $\mu_n$ on $X$ as $\mu_n (A) = \frac{\# A \cap Y_{\epsilon_n} }{\# Y_{\epsilon_n}}$. Now $\mu_n \to \mu$, weakly, independently of the choices of nets.

See the following post for the above construction: Does every compact metric space have a canonical probability measure?

This construction can not be regarded as canonical, since it depends on the sequence $\{\epsilon_n\}$ that is employed. (An example is given in the comments on the linked post.)

My question: Is the choice of sequence irrelevant if $X$ is a compact subset of a Banach space? If yes, compact subsets of Banach spaces may carry a canonical uniform measure.

Many thanks!


The answer is no since every compact metric space can be isometrically embedded into a Banach space.

  • $\begingroup$ What about compact subsets of a Hilbert space? There exists a simple example of a 4-element metric space that does not embed to a Hilbert space. $\endgroup$ – Taras Banakh Sep 30 '17 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.