TITLE: Interplay of Hausdorff metric and Lebesgue measure
QUESTION [7 upvotes]: 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.
Thanks.

REPLY [6 votes]: 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 $X_n$ is also a null-set for sufficiently large $n$. 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.