# Let $U$ be an open set of $R^n$, and let $f: U \to R^n$ be a smooth map. If $A \subset U$ is a measure zero, then $f(A)$ is of measure zero.

Let $U$ be an open set of $R^n$, and let $f: U \to R^n$ be a smooth map. If $A \subset U$ is a measure zero, then $f(A)$ is of measure zero.

Proof (Differential Topology - Guillemin and Pollack): $\textbf{We may assume that}$ $\overline{A}$ $\textbf{is compact and contained in}$ $U$, $\textbf{since}$ $A$ $\textbf{may be written as a countable union}$ $\textbf{of subsets with this property.}$ Let $W$ be an open neighborhood of A such that $\overline{W}$ is compact and contained in U. Since $\overline{W}$ is compact, there exists a constant $M$ such that $\mid f(x) - f(y) \mid < M \mid x- y \mid$ for all pairs $x,y \in \overline{W}$. It follows that there exists another constant $M'$ such that if $S$ is any cube in W, then f(S) is contained inside a cube $S'$ of volume less than $M' vol (S)$. Now show that the set A can be covered by a sequence of cubes $S_1, S_2,...$, each contained in $W$, with $\sum vol (S_i)$ less than any prescribed $\epsilon$. Thus we get a covering $S'_1, S'_2,...$ of f(A) by cubes with $\sum vol (S'_i) < M' \epsilon$. As $\epsilon$ is arbitrary, f(A) has measure zero. Q.E.D

How can we show that you can cover A by countably many sets $A_1, A_2, A_3...$ whose $\overline{A_i}$ are compact and contained in U. How do we write A as a countable union of subsets $A_1, A_2, ...$ whose closures are compact and contained in U? One choice is to let $A_i = A \cap \{x \in R^n: \mid\mid x \mid\mid < i\}$, but I'm not certain.

If $U = \mathbb {R}^n,$ this is simple. If $U\ne \mathbb R ^n,$ then $U^c$ is nonempty. Here, for $m=1,2,\dots,$ let $K_m= \{x\in U: d(x,U^c) \ge 1/m\}\cap \{|x|\le m\}.$ These are compact sets in $U$ whose union is $U.$ Then you can let $A_m= A \cap K_m.$
• Are you able to complete all the proof for this question? It is not clear why this is simple in the case $U = \mathbb {R}^n$ – user230283 Jul 10 '15 at 17:37
• In that case you just take $K_m = \{|x|\le m\}.$ – zhw. Jul 10 '15 at 17:52
• But I don't understand why you add the condition $\{x\in U: d(x,U^c) \ge 1/m\}$ – user230283 Jul 10 '15 at 19:09
• Try something simple like $U=$ the upper half plane in $\mathbb R ^2$ where you can see the $K_m$'s explicitly. – zhw. Jul 10 '15 at 19:16