Prove that a $C^1$ map $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ takes sets of measure $0$ to sets of measure $0$. Prove that a $C^1$ map $f:R^n \rightarrow R^n$ takes sets of measure $0$ to sets of measure $0$. That is to say, if $A$ is a measurable set and $m(A)=0$, then $m(f(A))=0$, which m is the Lebesgue measure of $R^n$.
How to prove it? And the next versions,
1.If $f:U$(is a nonempty open set of $R^n) \rightarrow R^n$ is $C^1$, then the conclusion is also right.
2.For $0<m<n$, $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ takes sets of measure $0$ to sets of measure $0$?
 A: Use the change of variables formula: for a measurable set $E$,
$$
\int_{f(E)} 1d\mu = \int_E |\det f'|\,d\mu
$$
Now, if $E$ is measure zero, note that we can write
$$
E = \bigcup_{k=1}^\infty \{x \in E: |\det f'(x)| \leq k\}
$$
Note that each set in this countable union is mapped to a set of measure zero.

Statement 3 is clearly true since we can think of $\Bbb R^m$ as a measure-$0$ subset of $\Bbb R^n$.
A: *

*Assume that $f:U\to\mathbb R^n$ is $\mathcal C^1$, where $U\subseteq\mathbb R^n$ is open. Let $A\subseteq U$ be a measure $0$ set.


The key point is that, being $\mathcal C^1$, the function $f$ is locally lipschitz (because its differential is continuous and hence locally bounded). It follows that the open set $U$ is the union of countably many open sets $V_i$ such that $f$ is lipschitz on $V_i$ for each $i$. Then $f(A)$ is the union of the sets $f(A\cap V_i)$, and it is enough to show that every such set is a measure $0$ set. So, we may assume from the beginning that $f$ is lipschitz on $U$. In what follows, we fix a constant $M$ such that $f$ is $M$-lipschitz on $U$, where (for convenience) we take the norm $\Vert\,\cdot\,\Vert_\infty$ on $\mathbb R^n$ (so that the balls are in fact cubes).
Let $\varepsilon >0$ be given. Since $A$ is a measure $0$ set, one can find a sequence of closed cubes $(C_k)_{k\in\mathbb N}$ such that $$A\subseteq \bigcup_{k=1}^\infty C_k\subseteq U\qquad{\rm and}\qquad \sum_{k=1}^\infty \vert C_k\vert<\varepsilon\, .$$ 
Since $f$ is $M$-lipschitz, each set $f(C_k)$ is contained in a cube $C'_k$ of side-length $M$ times the side-length of $C_k$. Then 
$$f(A)\subseteq\bigcup_{k=1}^\infty C'_k\qquad{\rm and}\qquad m(f(A))\leq\sum_{k=1}^\infty \vert C'_k\vert=M^n\sum_{k=1}^\infty \vert C_k\vert<M^n \,\varepsilon\, .$$
Since $\varepsilon>0$ is arbitrary, it follows that $f(A)$ is a measure $0$ set.


*I think what has to be proved here is that if $f:\mathbb R^m\to\mathbb R^n$ is $\mathcal C^1$ and $m<n$, then $f(\mathbb R^m)$ is a measure $0$ set. Since $\mathbb R^m$ is the union of countably many closed cubes, it is enough to show that $f(C)$ is a measure $0$ set for every cube $C\subseteq \mathbb R^m$. Le us denote by $c$ the side-length of $C$.


As before, the function $f$ is lipschitz on $C$, say $M$-lipschitz. Let $N$ be an arbitrary positive integer. Then $C$ is the union of $N^m$ cubes of side-length $\frac{c}{N}\cdot$. Since $f$ is $M$-lipschitz, it follows that $f(C)$ is contained in the union of $N^m$ cubes of side-length $\frac{Mc}{N}\cdot$ Therefore
$$m(f(C))\leq N^m\times \left(\frac{Mc}N\right)^n=\frac{K}{N^{n-m}}\, ,$$
where $K:=(Mc)^n$. Since $n>m$ and $N$ is arbitrary, this shows that $m(f(C))=0$.
Alternatively (as suggested by Omnomn...), you can argue as follows. Define $\widetilde f:\mathbb R^n\to\mathbb R^n$ by 
$$\widetilde f(x_1,\dots ,x_n):=f(x_1,\dots ,x_m)\, .$$
Then $\widetilde f$ is $\mathcal C^1$, and $f(\mathbb R^m)=\widetilde f(\mathbb R^m\times \{ 0\})$, where $0=(0,\dots ,0)\in\mathbb R^{n-m}$. Since $\mathbb R^m\times \{ 0\}$ has measure $0$ in $\mathbb R^n$, the result follows from the previous one.
