It is well known that if $K\_n$ are compact sets in $\mathbb{R}^n$ converging in Hausdorff distance to $K$ compact as well, then it does not follow that their Lebesgue measures converge (even if the Lebesgue measure of the $K\_n$ is constant).


Given this, I would like to ask the following question: If we assume that $K\_n$ does not only converge in HD to $K$ but also in measure $\lambda(K\_n) \rightarrow \lambda(K)$ and let $F$ be a Lipschitz function, does it then follow that $\lambda(F(K\_n)) \rightarrow \lambda(F(K))$? 


This question is related to the previous since Lipschitz continuity still preserves metric convergence of $F(K\_n)$ to $F(K)$, but now we have also the information that $\lambda(K\_n) \rightarrow \lambda(K)$. Does this allow us to deduce the claim?



Convergence in Lebesgue measure

Note the following lemma: under the given assumptions, we have $\lambda(K\_n \triangle K) \to 0$. Fix $\epsilon > 0$ and choose a $\delta$ so small that the $\delta$-thickening $K^\delta$ of $K$ has measure $\lambda(K^\delta) < \lambda(K) + \epsilon$. (This is possible since $K$ is compact and hence $\bigcap\_i K^{1/i} = K$.) Now if $n$ is so large that $K\_n \subset K^\delta$ (by Haudsdoff convergence) then $$\lambda(K\_n \setminus K) \le \lambda(K^\delta) - \lambda(K) < \epsilon$$
and if we further take $n$ large enough that $\lambda(K\_n) > \lambda(K) - \epsilon$, then we also get
$$\lambda(K \setminus K\_n) \le \lambda(K^\delta) - \lambda(K\_n) < 2 \epsilon$$
proving the lemma.


Now note that $F(K\_n) \triangle F(K) \subseteq F(K\_n \triangle K)$. As such, we have
$$\lambda(F(K\_n) \triangle F(K)) \le \lambda(F(K\_n \triangle K)) \le C\_F \lambda(K\_n \triangle K) \to 0$$
where $C\_F$ is the Lipschitz constant of $F$. In particular $\lambda(F(K\_n)) \to \lambda(F(K))$.


(Thanks to Iosif Pinelis for suggesting this argument which is much simpler than what was here before.)


