I'm trying to proof that given a diffeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$, the image of a set of measure zero, is also a set of measure zero.
I've seen several answers to similar questions here before, but unfortunately at the specific course I'm doing at the moment, we are required to talk about sets of measure zero using boxes, namely sets of the form $$[a_1,b_1]\times\dots\times[a_n,b_n]$$
So assuming my set is bounded for the moment, I know I can think of the set $E$ as a subset of some compact set containing it. So generally speaking, I know that $f$ must be Lipshitz for some constant $M$ so: $$\lVert f(\vec{x})-f(\vec{y})\rVert\leq\lVert \vec{x}-\vec{y}\rVert$$
Now, since $E$ is of measure-zero, our definition says that given some $\varepsilon>0$ we have a countable collection $\{A_i\}_{i=1}^{\infty}$ whose union covers $A$ and the sum of the volumes is less than $\varepsilon$.
I was trying to use that to show that $f(E\cap A_i)$ is contained is a box with a volume bounded by $M{\rm Vol}\left(A_i\right)$, and this will finish the proof, but I'm somehow stuck. I can show the set is bounded inside some ball of radius $\max\limits_{i=1,\dots,n}b_i-a_i$ but I can't make it proportional to the volume.
Any help would be greatly appreciated.