For a Polish space $X$, let $K(X)$ be the set of compact subsets of $X$. Given the topology with basis $\{K\in K(X):K\subset U_0, K\cap U_1\neq\emptyset,\ldots,K\cap U_n\neq\emptyset\}$ for open sets $U_i$, $K(X)$ is also Polish. A compatible metric is the Hausdorff metric

I was wondering if in cases where there is a natural measure on $X$, say Lebesgue measure on $[0,1]$ or the uniform measure on $2^{\omega}$, if there is a natural measure on $K(X)$? By natural I suppose I mean that it can somehow be described in terms of the measure on $X$.


In some cases a "natural" measure may be a Hausdorff measure (if it exists) that is positive and finite. So we want not just a Polish topology, but a specific metric. Given a metric for $X$ itself, use the Hausdorff metric on $\mathcal{K}(X)$.

In some cases ($X$ countable, with box dimension zero) it may happen that $\mathcal{K}(X)$ has positive, finite Hausdorff dimension. Then use the Hausdorff measure in that dimension.

But typically $\mathcal{K}(X)$ has infinite Hausdorff dimension. There still may be a gauge function $\phi$ so that the Hausdorff measure $\mathcal{H}^\phi$ is positive and finite on $\mathcal{K}(X)$. But it may happen that you always get measure zero or infinity. You can still try to classify $\phi$ into these two classes.

Given a good gauge function for $X$, how is that related to a good gauge function for $\mathcal{K}(X)$?

My student Mark McClure worked out a few of these in his thesis. Also see his references:

McClure, Mark, "The Hausdorff dimension of the hyperspace of compact sets". Real Anal. Exchange 22 (1996/97), no. 2, 611–625.

McClure, Mark, "Entropy dimensions of the hyperspace of compact sets". Real Anal. Exchange 21 (1995/96), no. 1, 194–202.


You know by the measurable isomorphism's theorem that between Polish probability spaces without atoms there is a measurable bijection that takes one probability measure into the other. Therefore, any probability measure without atoms you take in $K(X)$ can be somehow described in terms of the measure on $X.$


$X$ embeds into $K(X)$ (with the topology you have defined) by $x\mapsto\{x\}$, so any measure you have on $K(X)$ restricts to $\{\{x\}: x\in X\} \subseteq K(X)$ which is a homeomorphic copy of $X$.

  • $\begingroup$ I think the question was not how to get a measure on $X$ if you have a measure on $K(X)$ but the other way around. $\endgroup$ – Tom Nov 8 '16 at 13:26

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