Prove that the range set of a Lipchitz function is zero content A half-open rectangle in $\mathbb{R}^n$ is of the form $P=(a_1,b_1] \times \cdots \times (a_n,b_n]$.
Denote the collection of all half-open rectangles by $P'$
Define a function $Vol_0 :P' \rightarrow [0,\infty)$ by $Vol_0 (P)=(b_{1}-a_{1})\cdots (b_{n}-a_{n})$
We say a set $A \subseteq \mathbb{R}^n$ is zero content if for every $\epsilon >0$, there exist a finite number of half-open rectangles $P_1, \ldots,P_l$ (for some $l \in \mathbb{N}$) such that $P_1 \cup \cdots \cup P_l \supseteq A$ and $\sum_{i=1}^{l} Vol_0 (P_i) < \epsilon$
We say a function f is Lipshitz on a set $B \subseteq \mathbb{R}^n$ if there exists a number $c \geq 0$ such that $$||f(\vec{x_1})-f(\vec{x_2})|| \leq c ||\vec{x_1}-\vec{x_2}|| \ \forall \vec{x_1},\vec{x_2} \in B$$
Question: Let $X=[0,1] \times \cdots \times [0,1] \subseteq \mathbb{R}^{n-1}$. Let $f:X \rightarrow \mathbb{R}^n$ be Lipschitz on X. Prove that the set Range(f)=f(X) is zero content.
My solution is as follows :
Write $f(\vec{x})=(f^{1}(\vec{x}),\ldots,f^{n}(\vec{x}))$
Then for every $\epsilon >0$,choose $n<\frac{\epsilon}{4}$ and  we construct the half-open rectangles $P_1, \ldots,P_{l}$ (for some $l \in \mathbb{N}$ by $P_i=(a_{i1},b_{i1}] \times \cdots \times (a_{in},b_{in}]$ for $1 \leq i \leq l$ with $a_{ij}=f^{j}(\vec{x})-\frac{1}{i}$ and $b_{ij}=f^{j}(\vec{x})+\frac{1}{i}$ for $1 \leq j \leq n$
Now we have $P_1 \cup \cdots \cup P_l \supseteq f(X)$ and $\sum_{i=1}^{l} Vol_0 (P_i) =\sum_{i=1}^{l} \sum_{j=1}^n \frac{2}{i} =2n \cdot \sum_{i=1}^l \frac {1}{i} \leq 2n \cdot 2=4n<4 \cdot \frac{\epsilon}{4} = \epsilon$
However, my solution is obviously wrong, because i did not use the fact that f is Lipschitz on $X=[0,1] \times \cdots \times [0,1] \subseteq \mathbb{R}^{n-1}$
What's missing in my solution?
 A: Hmm, where to start?


*

*You don't get to choose $n$. That has to be the dimension of the codomain of $f$, which is given to you when someone wants to apply the result.

*On the other hand, $l$ seems to come out of nowhere. But the end of your calculation depends on it being $n$ that is chosen rather than $l$.

*Even so, you're choosing $n$ (or perhaps $l$) to be less than $\varepsilon/4$. If $\varepsilon$ is small, then that means that $n$ (or $l$) have to be $0$, in which case it seems that you have no rectangles, or no factors in the product you make the rectangles with.

*Your $P_i$s seem to depend on a particular choice of $\vec x$, but no $\vec x$ has been chosen. The only use of $\vec x$ so far is as the independent (dummy) variable in the definition of $f$.

*What makes you think the union of the $P_i$s is a superset of $f(X)$? You have constructed the boxes to have side length at most $2$, but who says the range of $f$ stays within such a box? The points of $X$ stay within a box of side length $1$, but you don't know whether they stay together when you apply $f$ to them.

*The volume of your $P_i$ is $\prod_{j=1}^n \frac2i$, not $\sum_{j=1}^n \frac2i$.

*You seem to assume that $\sum_{i=1}^l \frac1i \le 2$ for all $l$. That is definitely not true -- the sum is the harmonic number $H_l$, which is greater than $2$ as soon as $l\ge 4$.
