Spivak "Differential Geometry "volume 1 page 41 I want to ask question related to Lemma 6 from Spivak "Differential Geometry Volume 1" page 41. The lemma states: If $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is $C^{1}$ and $A\subset\mathbb{R}^{n}$ has measure 0, then $f(A)$ has measure 0. 
The author claims that we can assume $A$ to be contained in a compact set $C$, and because $|f(x)-f(y)|\leq n^{2}K|x-y|$ for all x,y in C, f takes rectangles of diameter d into sets of diameter $\leq n^{2}Kd$. Then, Spivak wrote" This clearly implies $f(A)$ has zero measure if $A$ does". 
Could anyone enlighten me why he can straightaway claimed that $f(A)$ has zero measure? To my knowledge so far, set which has zero measure might not imply that it has zero diameter (for instance, set of all rational numbers between 0 and 1).
 A: We can write $\mathbb{R}^n$ as the union of countably many compact boxes:
$$
\mathbb{R}^n=\bigcup_{m=(m_1,\dots,m_n)\in\mathbb{Z}^n}C_m,
$$
where
$$
C_m=\prod_{i=1}^n[m_i,m_i+1].
$$
Then if $A\subset\mathbb{R}^n$ has measure zero, $A_m:= A\cap C_m$ has measure zero as well. Spivak's argument, as outlined in your post, shows that $f(A_m)$ has measure zero for all $m\in\mathbb{Z}^n$. Since we have 
$$
f(A)=f\left(\bigcup_{m\in\mathbb{Z}^m}A_m\right)=\bigcup_{m\in\mathbb{Z}^m}f(A_m)
$$
because
$$
A=A\cap\mathbb{R}^n=A\cap \bigcup_{m\in\mathbb{Z}^n}C_m=\bigcup_{m\in\mathbb{Z}^n} (A\cap C_m)=\bigcup_{m\in\mathbb{Z}^n}A_m,
$$
we conclude that $f(A)$ has measure zero because a countable union of sets of measure zero has measure zero. This is why Spivak said that one can assume $A$ is contained in a compact set.
Here is why $A_m$ has measure zero: Our set $A_m$ is contained in the compact set $C_m$. On the dilation $2\cdot C_m$, we have that
$$
\|f(x)-f(y)\|\leq M\|x-y\|
$$
for some constant $M>0$ since $2\cdot C_m$ is still compact and $f$ was assumed to be $C^1$. As a consequence of this, if $B_r(x)\subset 2\cdot C_m$ is the ball of radius $r$ centered at $x$ in $\mathbb{R}^n$, then if $f(y)\in f(B_r(x))$, we have
$$
\|f(y)-f(x)\|\leq M\|x-y\|<Mr.
$$
Thus, $f(B_r(x))$ is contained in the ball of radius $Mr$ centered at $f(x)$, $B_{Mr}(f(x))\subset\mathbb{R}^n$. We have $m(B_{Mr}(f(x)))=M'm(B_r(x))$, where $M'>0$ is some constant depending only on $M$ and the dimension $n$.
Let $\delta>0$. Since $m(A_m)=0$, we can cover $A_m$ by a countable collection of balls $(B_k)_{k=1}^\infty$ such that 
$$
\sum_{k=1}^\infty m(B_k)<\delta.
$$
Taking $\delta$ small enough, we may assume that $B_k\subset 2\cdot C_m$ for all $k\in\mathbb{N}$. So, for each $k\in\mathbb{N}$, by the discussion above we have that $f(B_k)$ is contained in some open ball $B_k'$ of measure at most $M'm(B_k)$, so that
$$
\sum_{k=1}^\infty m(B_k')\leq M'\sum_{k=1}^\infty m(B_k)<M'\delta,
$$
and since the $B_k$'s cover $A_m$, the $B_k'$'s cover $f(A_m)$. Thus, $f(A_m)$ can be covered by countably many open balls whose measures sum to $M'\delta$. We may take $\delta$ to be arbitrarily small, showing that $f(A_m)$ has measure zero.
