Interplay of Hausdorff metric and Lebesgue measure 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.
 A: 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 $L^1$. 
A: 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.
