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I have a question linked to Interplay of Hausdorff metric and Lebesgue measure. Let us denote as $\mathcal K(\mathbb R^n)$ the space of compact subsets of $\mathbb R^n$ endowed with the Hausdorff metric $\rho$ and let $\lambda$ be the $n$-dimensional Lebesgue measure on $\mathbb R^n$. I want to know if there are (sufficient) conditions under which the measure $\lambda$ is continuous w.r.t. $\rho$, that is $$ \lim_{k\rightarrow\infty}\rho(K, K_k)=0\qquad\Rightarrow\qquad \lim_{k\rightarrow\infty}\lambda(K_k)=\lambda(K).\qquad(\star) $$ I tried to search it in the books Fractal geometry by Kenneth Falconer and Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco and Pallara but I did not find anything. In the second book it is written that, in the case $n=2$, the Hausdorff measure (which is a rescaling of the usual $\lambda$ on $\mathbb R^n$) is lower-semicontinuous w.r.t. the Hausdorff metric along sequences satisfying a suitable uniform concentration property, but this is not what I am looking for.

Some help? Do you have some references?

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    $\begingroup$ $K$ has to be zero measure since finite sets are dense. Is there a zero measure compact that is a limit of $>\epsilon$ measure compacts? $\endgroup$
    – Ville Salo
    Commented Jun 25, 2020 at 14:42
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    $\begingroup$ If $K_k$ converges to $K$, then for any $\varepsilon>0$, $K_k$ is eventually included in $K+B(0,\varepsilon)$. Since $\lambda(K+B(0,\varepsilon))\to0$ as $\varepsilon\to0$ (Lebesgue dominated convergence), the Lebesgue measure is semicontinuous (I want to say upper?). $\endgroup$
    – Pierre PC
    Commented Jun 25, 2020 at 14:46
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    $\begingroup$ Proof by Socratic method $\endgroup$
    – Ville Salo
    Commented Jun 25, 2020 at 14:49
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    $\begingroup$ Conjecture (that I think is not difficult to prove with the above comments): for all neighbourhood $U$ for some fixed $K$, the set $\lambda_*(U)=\lbrace \lambda(K'),\ K'\in U\rbrace$ contains $[0,\lambda(K)]$; more precisely, the intersection of all $\lambda_*(U)$ where $U$ ranges over all neighbourhoods of $K$ is precisely $[0,\lambda(K)]$. I think I can write that down if it is of interest. $\endgroup$
    – Pierre PC
    Commented Jun 25, 2020 at 14:54
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    $\begingroup$ @VilleSalo Yes, precisely the interval $[0,\lambda(K)]$. I must say though, the answers on math.SE are rather detailed already (I just checked it), so I'm not sure what the OP has in mind. $\endgroup$
    – Pierre PC
    Commented Jun 25, 2020 at 15:11

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