Consider the space $\mathcal{K}(\mathbb{R}^{n})$ of compact subsets of $\mathbb{R}^{n}$ endowed with the Hausdorff metric $\rho$, and let $\lambda$ denote the $n$-dimensional Lebesgue measure on $\mathbb{R}^{n}$. Now, I understand that $\lambda$ is not continuous with respect to $\rho$, i.e. that $[ \lim_{k \to \infty} \rho(K, K_{k}) = 0 ] \not \Rightarrow [ \lambda(K) = \lim_{k \to \infty} \lambda(K_{k}) ]$. But my question is: Suppose $\lambda(K_{i}) = \lambda(K_{j}) = C$ for every $i, j \in \mathbb{N}$, i.e. the sequence $k \mapsto \lambda(K_{k})$ is constant; does this imply that $\lambda(\lim_{k \to \infty} K_{k}) = C$? If not, I'm interested to see a counterexample.



2 Answers 2


Counterexample. In one dimension, let $K_k=\{0,1/k,2/k,3/k,\dots,1\}$. Each $K_k$ has zero measure. The limit of $K_k$ in the Hausdorff metric is the interval $[0,1]$, of measure $1$.

You can also attach the same interval to the sets $K_k$ to make their measure positive.

The above counterexample could be ruled out by requiring $K_k = \overline{\operatorname{int} K_k}$. But then there's a different one: $K_k = \bigcup_{j=1}^k [(2j-1)/2k,j/k] $ where each set has measure $1/2$ but the limit is again $[0,1]$.

I don't think you can benefit from Hausdorff-metric convergence here at all. I suggest looking at sufficient conditions for $L^1$ convergence, so that you can get $\chi_{K_k}\to \chi_K$ in $L^1$.

  • 2
    $\begingroup$ Interesting. Is there any useful hypothesis you could add to get that $\lambda( \lim_{k} K_{k}) = C$? $\endgroup$
    – AJY
    Jun 26, 2015 at 19:35
  • 1
    $\begingroup$ Expanded my answer. $\endgroup$
    – user147263
    Jun 26, 2015 at 19:46
  • $\begingroup$ I think that a potential useful restriction is the case where $K_k$ and $K$ are convex. $\endgroup$ Jul 4, 2020 at 22:19

As pointed out, the answer is negative. However, Lebesgue measure is upper semi-continuous with respect to the Hausdorff metric. That is, the measure can only increase in the limit, not decrease.

Indeed, suppose $X_n\to X$ in the Hausdorff metric, and $C:= \lambda(X)<\infty$. Let $U\supset X$ be an open set of measure at most $C+\varepsilon$. Then, for sufficiently large $n$, $X_n\subset U$, and hence $\lambda(X_n)\leq C+\varepsilon$. So indeed $$ \limsup_{n\to\infty} \lambda(X_n) \leq \lambda(X),$$ as claimed.

In particular, if $X$ is a null-set, then the measure of $X_n$ tends to zero as $n\to\infty$. As pointed out in the accepted answer, this is the only case where the answer to your question is positive, since any compact set is the Hausdorff limit of a sequence of finite sets.

  • $\begingroup$ Is the upper semi-continuity still holding if the compactness of $X_n$ and $X$ is dropped ? $\endgroup$
    – Pohoua
    Apr 21, 2021 at 17:25
  • $\begingroup$ @Pohoua What do you mean by "dropping compactness"? I think your sets need to be at least closed if Hausdorff convergence is to make any sense. I don't think you can say anything about measure (with respect to usual Lebesgue measure and Euclidean distance here) since you may have a set of zero measure that is denser and denser as you get close to infinity, so that every point of size at least $R$ has distance at most $\varepsilon(R)$ to the set, where $\varepsilon(R)\to 0$ as $R\to\infty$. $\endgroup$ Feb 11 at 22:28

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