Existence of the compact m-cube sequence.

Question

I have a problem connected with the proof of following lemma:

If we have $Z \subset \Bbb R^n$, $m \ge 1$, $\mathcal L^m(Z)=0$ (Lebesgue measure) and $f: Z \rightarrow \Bbb R^m$ locally lipschitz,
then $\mathcal L^m(f(Z)) = 0$.

Proof:
from the second axiom of countability we can assume, that $f$ is lipschitz, $Lip(f)=M< \infty$. Lets take $\epsilon > 0$.
Now we got to the point: in the proof it's assumed, that there exists $\{I_n\}_{n\in\Bbb N}$ such that $I_n$ is a compact m-dimensional cube with side length $\delta _n$ for $n\in \Bbb N$ such that $Z \subset \cup_{n\in\Bbb N} I_n$ and $\sum_{n\in\Bbb N } (\delta _n)^m < \epsilon $

Please help me with showing the existance of such $\{I_n\}_{n\in\Bbb N}$.

(!) Edit: I am interested in knowing, what sequence will satisfy the existence.

Answer 1

I'm assuming that ${\cal L}^m$ means "Lebesgue measure". The existence of such a sequence $(I_n)_{n\geq1}$ of cubes for each $\epsilon>0$ is the exact definition of a set of Lebesgue measure zero; see here:

http://en.wikipedia.org/wiki/Measure-zero_set

Therefore, if it is assumed that ${\cal L}^m(Z)=0$ then the existence of such a sequence is implied.