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If one considers the metric space $\mathcal{K}$ of all compact subsets of $\mathbb{R}^2$ endowed with the Hausdorff distance (that is $\Delta(A,B)=\inf \{ \delta: A\subset B^{\delta},B \subset A^{\delta} \}$). Is it true (and how can it be shown in that case) that the function $\mu: \mathcal{K} \to \mathbb{R}^2$ ($\mu$ is the 2-dimensional lebesgue measure) defined by $A \mapsto \mu{(A)}$ is continuous w.r.r the metric $\Delta$?

user202542
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  • I'm skeptical. My preliminary investigations suggest that fractal shapes might cause this to fail. It seems that adding a $\delta$-ball to a shape seems to add area proportional to the length of the perimeter. Have you tried looking at, say, Koch's Snowflake? – Theo Bendit Dec 05 '17 at 13:47
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    It is lower semicontinuous but not continuous. https://math.stackexchange.com/questions/1339102/interplay-of-hausdorff-metric-and-lebesgue-measure?rq=1 – Moishe Kohan Dec 05 '17 at 13:51
  • @Theo Bendt: As you suspected, this fails. In fact, it fails in a major way --- the collection of finite sets is dense in $\mathcal{K}.$ See Finite sets are dense with respect to Hausdorff distance. Also, the sets of Lebesgue measure zero (even the sets of Hausdorff dimension zero) are "superdense" in the sense they form a co-meager subset of $\mathcal{K}.$ – Dave L. Renfro Dec 05 '17 at 16:44
  • Thanks for your replies! Lower semicontinuous in the sense that if $A_1 \subset A_2 \subset \dots \subset A_n \subset \dots A$ then $\mu(A_n) \to \mu(A)$? How does one show this? – user202542 Dec 06 '17 at 09:27
  • @MoisheCohen: I think you mean upper semicontinuous? Suppose $A_k \to A$ in Hausdorff distance. Choose a thickening $A^\delta$ of $A$ whose measure is only slightly larger than that of $A$ (this is possible since $A$ is compact and hence $A = \bigcap_n A^{1/n})$, so by countable additivity we have $\mu(A) = \lim \mu(A^{1/n})$). Now almost all the $A_k$ are contained in $A^\delta$, so we conclude $\limsup \mu(A_k) \le \mu(A)$, which is upper semicontinuity. Lower semicontinuity fails as the usual counterexamples show. – Nate Eldredge Jan 08 '18 at 05:09
  • @NateEldredge Right, this is what I had in mind. – Moishe Kohan Jan 08 '18 at 17:39

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