Smooth image of a null set has measure zero Currently I'm looking at the proof of Sard's theorem given in John Milnor's "Topology from the differentiable viewpoint". I'm a bit confused about a remark on the case where the dimension of the target space is strictly larger than the dimension of the initial space. 
Let $f: U \rightarrow \mathbb{R}^n$ be a smooth map, where $U \subset \mathbb{R}^m$ is open and $m<n$. In this case it is clear that set of critical points of $f$ is just $U$ and here, Sard's theorem simply states that $f(U)$ has measure zero. Milnor says that this is an obvious fact, however, as I'm a bit rusty in measure theory, I wasn't able to proof this.
I also wonder whether the same is true if $f$ is just continuous.
Let me restate the question:
Let $f: U \rightarrow \mathbb{R}^n$ be smooth/continuous, where $U \subset \mathbb{R}^m$ is open and $m<n$. Does $f(U)$ have Lebesgue measure zero?
I found a few partial answers here and here. But while the first post is not general enough the second one is too general and I think in the case of a smooth map the proof should be quite elementary.  
Thank you for any help
 A: The comment by Thomas is exactly right (the locally-Lipschitz property is what one normally uses here), but I'll add an argument for the smooth case. By exhausting $U$ with compactly contained subdomains, we reduce to the case when the derivative $Df$ is bounded, say $\|Df\|\le M$. Fix $\epsilon>0$. For each point $x\in U$ there is $r=r_x>0$ such that $$|f(y)-Df(x)(y-x)|\le \epsilon |y-x|\quad\text{ whenever } \ |y-x|<r$$ The image of the ball $B(x,r)$ under the linear map $Df(x)$ is contained in the intersection of an $m$-dimensional affine subspace with $B(f(x),Mr)$. By (1), the image of $B(x,r)$ under $f$ is contained in $(\epsilon  r)$-neighborhood of said intersection, which is (more or less)  its product with an $(n-m)$-dimensional ball of radius $\epsilon$.  Therefore, the $n$-dimensional measure of $f(B(x,r))$ is at most $Cr^m\epsilon^{n-m}$. 
Applying Vitali's lemma to the cover of $U$ by $B(x,r_x/5)$ we get a  cover of $U$ by $B(x_i,r_{x_i})$ such that $B(x_i,r_{x_i}/5)$ are disjoint. Hence, the Lebesgue measure of $f(U)$ is at most 
$$
C\epsilon^{n-m} \sum_i r_i^m \le   C  \epsilon^{n-m} 5^m \operatorname{vol}_m U
$$
Since $\epsilon>0$ was arbitrary, the conclusion follows.
