# Convergence in Lebesgue measure

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?

• There are counter examples if $F$ is allowed to change dimension e.g. $F(x,y)=x$. – Anthony Quas Jan 8 '18 at 2:19

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 $\lambda(F(K_n)) \to \lambda(F(K))$.
• I wanted to give a similar answer, but you did it first! Nice answer. However, it looks like the change of variables formula you referred to works only for bi-Lipschitz maps. However, it is easy to see that for any Lipschitz map $F$ there is a real number $C_F$ such that $\lambda(F(A))\le C_F\lambda(A)$ for all Lebesgue-measurable sets $A$ (cover $A$ tightly enough by countably many boxes, say). – Iosif Pinelis Jan 8 '18 at 5:37