Equivalent definition of measure zero in a Manifold.

Question

In Munkres's analysis on manifold, it defines:
Let $M$ be a compact k-manifold in $\mathbb{R^n}$, of class $C^r$. A subset $D$ of $M$ said to have measure zero in $M$ if it can be covered by countably many coordinate patches $\alpha_i: U_i\rightarrow V_i$ such that the set $D_i ={\alpha_i}^{-1}(D\cap V_1)$ has measure zero in $\mathbb{R^k}$.

And it also give a equivalent definition:
For any coordinate patch $\alpha:U\rightarrow V$ on $M$, the set $\alpha^{-1}(D\cap V)$ have measure zero in $\mathbb{R}^k$.

And the proof is :
To verify this fact, it suffices to show that $\alpha^{-1}(D\cap V\cap V_i)$ has measure zero for each $i$. And this follows from the fact the set $\alpha_{i}^{-1}(D\cap V\cap V_i)$ has measure zero because it is a subset of $D_i$, and that $\alpha^{-1}\circ\alpha_i$ is of class $C^r$.

I cannot follow the last line of this proof. Is it saying since the composition $\alpha^{-1}\circ\alpha_i\circ\alpha_{i}^{-1}(D\cap V\cap V_i)=\alpha^{-1}(D\cap V\cap V_i)$, and $\alpha^{-1}\circ\alpha_i$ is continuous, so it keeps the measure zero of $\alpha_{i}^{-1}(D\cap V\cap V_i)$? But continuous functions do not preserve measure zero. Where is wrong? Thanks!

Answer 1

Continuous functions don't preserve measure zero. Continuously differentiable functions do. We need $r>0$ to rule out examples in the vein of the "devil's staircase", but once we're there we have that the measure of $f(X)$ is at most $\sup_{x\in X}\|f'(x)\|^k$ times the measure of $X$.

And yes, we need the additional (presumably, appearing somewhere earlier in the text) assumption that $r>0$. Without that, consider the line as a topological ($C^0$) $1$-manifold. There are homeomorphisms that send measure-zero sets like the Cantor set to sets of positive measure. With the concept of "measure zero" not invariant under homeomorphisms, we just can't define it on this manifold.