I'm reading proof that set with $\lambda_n$ measure zero have also $\mathscr{H}^n$ (Hausdorff) measure zero. And there is claim:
If $n\geq 2$, for every given $\epsilon,\delta > 0$ and Borel $\lambda_n$-null set $N \subset \mathbb{R}^n$ we cover N with sequence of cubes $(Q_i)$ ($N \subset \bigcup Q_i$) such that
$\texttt{diam } Q_i < \delta$ and $\sum \texttt{vol }Q_i < \epsilon$.
There is no proof of that and it looks super unintuitive.
I convinced myself that it may be true with simple example. $\lambda_1$ measure of $Q$ (rationals) is zero, and we can take sequence $(x_i)$ of them and cover them with sequence of balls $B_i=B(x_i, \frac{1}{2^i})$ for $i\in\mathbb{N}_+$.
Of course $Q \subset \bigcup B_i$ and $\sum\texttt{vol }B_i = 2$.
(Scaling to meet the requirements of $\epsilon$ and $\delta$ is easy.)
But I still don't see how extend this intuition to all Borel $\lambda_n$-null sets or how to prove it. Could you provide such an intuition and give me some hints how to prove it?
Best Answer
One can construct the Lebesgue-measure via the restriction of the outer measure$$\eta \colon A \mapsto \inf \left\{ \sum_{j=1}^n \text{vol}(Q_j) \vert A \subset \bigcup_j Q_j , Q_j \in \mathcal{D} \right\} \\ \text{ where } \mathcal{D}= \{ Q \subset \mathbb{R}^n : Q = [a_1 , b_1] \times \ldots \times [a_n , b_n] \}$$to the Borel-$\sigma$-Algebra $\mathcal{B}$. (By $\text{vol}$ I mean the natural volume of a cube: $\text{vol} (Q) = \prod_j (b_j-a_j)$ for a cube $Q = [a_1 , b_1] \times \ldots \times [a_n , b_n]$.)$$$$ So for any Lebesgue-null set $N \in \mathcal{B}$ the above infimum is zero. That means, by definition of the Lebesgue-measure, you get for $\varepsilon >0$ a covering$$N \subset \bigcup_{j=1}^n Q_j^\varepsilon$$with $\sum_{j=1}^n \text{vol} Q_j^\varepsilon < \varepsilon$. By splitting those $Q_j^\varepsilon$ into smaller cubes with volume $< \delta$ for a given $\delta$, you get what is claimed in your proof.