Lebesgue measure of $f(crit(f))$ is $0$ I have some problems to show this case of Sard's Theorem
Let $f:\mathbb R^2\to \mathbb R^2$ be $C^1$. Let $$
Crit(f)=\{x \in \mathbb R^2 ~:~ \det df_x=0\}
$$
then the Lebesgue measure of $f(Crit(f))$ is $0$.
Can you help me?
 A: Here is a sketch: Since $\mathbb R^2$ is a countable union of squares of unit side lengths, it suffices to prove the result on the unit square $I^2=I\times I$. 
Let $\epsilon>0.$ Define $F(x,y)=\frac{f(y)-f(x)-f'(x)(y-x)}{\|y-x\|}$ unless $x\neq y$, in which case set $F(x,x)=0.$ 
For any integer $n,$ we may subdivide $I^2$ into square blocks of side length $1/n.$ Since $f$ is $C^1$, there is an integer $N$ for which $\|F(x,y)\|<\epsilon $ on each such block of side length $1/N.$
Now suppose $c\in I^2$ is a critical point for $f.$ Then $c$ lies in one of the blocks and, for any other point $y$ in that block, we have 
$\frac{\|f(y)-f(c)-f'(c)(y-c)\|}{\|y-c\|}=\frac{\|f(y)-f(c)\|}{\|y-c\|}<\epsilon\Rightarrow \|f(y)-f(c)\|<\epsilon \|y-c\|\le \epsilon\cdot\frac{\sqrt 2}{N}$.  
An application of the triangle inequality now shows that if $y,z$ are $any$ two points in the block, then 
$\|f(y)-f(z)\|<\epsilon \|y-z\|\le 2\epsilon\cdot\frac{\sqrt 2}{N}.$
We conclude that the image under f of the critical points which lie in $I^2$ is contained in a union of blocks of total area $2\sqrt 2\epsilon,$ and therefore has measure zero.
