# Prove that Sets of Measure Zero Map to Sets of Measure Zero

Here is a lemma that I must prove as part of my midterm exam tomorrow:

Suppose $R\subset \mathbb{R}^n$ is a rectangle, $U\supset \overline R$, and $N\subset \overline R$ has measure zero. Let $\phi: U\to > \mathbb{R}^n$ be $C^1(U)$ i.e. its first derivative is continuous. Then $\phi(N)$ has measure zero.

Here is the proof that our teacher provided us in class. I bolded the part that I disagree with/do not understand.

$\phi \in C^1(U)$, $\overline R$ is compact, so $D\phi$ is bounded, i.e. $|D\phi (x)|\le M\ \forall x\in \overline R$.

So, $\forall x,y\in \overline R$, $\| \phi (x)-\phi (y)\| \le M*\| x-y\|$

Given $\epsilon >0$, let ${R_i}$ be a countable set of rectangles that cover N, with $\sum_{i=1}^{\infty}v(R_i)<\epsilon$.

If $x,y\in R_i$, then $\phi(x)$ and $\phi(y)$ are contained in a rectangle $R_i'$ of diameter $M*diam(R_i)$.

The volume of this rectangle is $(M*diam(R_i))^n\le M^n*v(R_i)$, so $\phi (N)$ is covered by rectangles $R_i'$ s.t. $\sum_{i=1}^{\infty}v(R_i')\le > \sum_{i=1}^{\infty}M^n*v(R_i)<M^n*\epsilon$.

Our given definition of the volume of a rectangle $R\subset \mathbb{R}^n$, with $R=[a_1,b_1]\times ...\times [a_n,b_n]$ is $\prod_{i=1}^n(b_i-a_i)$.

Our given definition of the diameter of a rectangle is $\sup_{x,y\in R}{\|y-x\|}$.

As far as I can tell, the volume of $R_i'$ should be $\le (M*diam(R_i))^n$, and $(M*diam(R_i))^n$ should be $\ge M^n*v(R_i)$. What's gone wrong? Can someone provide an alternative proof? It's getting really late and I need to get to bed, so I would really appreciate help ASAP. Thanks!

Edit: I think I might have come a solution:

$v(R_i)=\prod (a_i-b_i)$.

Thus $v(R_i')\le \prod (\phi (a_i)-\phi (b_i))\le \prod M*(a_i-b_i)=M^n*v(R_i)<M^n*\epsilon$

Thanks for explaining what you have tried! I agree with you in that I think there is an error here. What the bound should be, I believe, is $(M\cdot \text{diam}(R_i))^n\leq CM^n\cdot v(R_i)$ where $C$ is some constant independent of $R_i$ (i.e., independent of the rectangle).
In other words, we would like to find a constant $C$ (independent of $R$ in the following inequality) such that $(\text{diam}(R))^n\leq Cv(R)$ for all rectangles $R$. Can you think of what $C$ should be? It could possibly depend on the dimension, $n$.