Question About Lipschitz Maps and Measure Zero I'm having difficulty with a problem involving measure.

Suppose $F : B^k \rightarrow B^n$ is a Lipschitz map from the unit ball in $\mathbb{R}^k$ to the unit ball in $\mathbb{R}^n$. If $k \lt n$, show that the image of $F$ has measure zero in $B^n$.

Starting from the fact that $F$ is a Lipschitz map, I figured out that for all $\epsilon \gt 0, \forall F(x), F(y) \in B^n$, if $F(x), F(y) \in B_\epsilon$ in $B^n$, then $x, y \in B_{\epsilon/C}$ in $B^k$ where $C$ is the Lipschitz constant.
From here I'm thinking that one can cover the $k$ unit ball with some $n$-balls and then transfer those $n$-balls to $B^n$ where they cover image $(F)$. Then reducing epsilon reduces the volume of the $n$-balls faster than their number increases so the volume goes to zero.
I would appreciate any help or tips. Thank you.
 A: Maybe you can use the fact that a Lipschitz map is a.e. differentiable, say outside of a set $S$ with $m(S)=0$ , and then apply Sard's theorem (wherever differentiability holds, i.e., in $B^k-S$ ), so that every point in $\mathbb R^k \cap B^k$ is a critical point, so that $f(B^k -S)$ has measure zero in $B^n$. So all we need to deal with is the image of S under F.
Or, more easily,for this last, use that a Lipschitz function is absolutely continuous, so that , since S has measure zero, and a.continuity takes measure zero to measure zero, we have $m(f(S))=0$ , and you're done.
A: Considering $B^k$ as a subset of $B^n$, you can extend $F$ to a map $G:B^n\to B^n$ such that $G(x_1,\ldots,x_n)=F(x_1,\ldots,x_k)$.  Note that $G$ is Lipschitz with the same Lipschitz constant as $F$.  The measure of $B^k$ in $B^n$ is $0$, so given $\varepsilon>0$ there is a sequence of balls $\{B(\mathbf{x}_i,r_i)\}_i$ in $B^n$ such that $B^k\subseteq \cup_i B(\mathbf{x}_i,r_i)$ and $\sum_i m(B(\mathbf{x}_i,r_i))<\varepsilon$.  
Let $C>0$ be the Lipschitz constant of $F$, hence of $G$.  Then $G(B(\mathbf{x}_i,r_i))$ has diameter at most $2Cr_i$, and therefore it fits inside a ball of radius $2Cr_i$.  This implies that $m(G(B(\mathbf{x}_i,r_i))\leq (2C)^nm(B(\mathbf{x}_i,r_i)$.  Since 
$$F(B^k)=G(B^k)\subseteq G(\cup_iB(\mathbf{x}_i,r_i))=\cup_i G(B(\mathbf{x}_i,r_i)),$$
this implies that $m(F(B^k))<(2C)^n\varepsilon$. Since $C$ and $n$ are constants and $\varepsilon$ was arbitrary, this implies that $m(F(B^k))=0$.

Regarding your remarks, note that $d(F(x),F(y))<\varepsilon$ does not imply that $d(x,y)<\dfrac{\varepsilon}{C}$.  (The converse holds.)  For example, consider a constant map.  
