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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!

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

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  • $\begingroup$ 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

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