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?