Whitney decomposition

Remarks

The cubes can be chosen dyadic, that is, of the form $2^{-j}([0, 1]^n + m)$ with $j \in \Z$ and $m \in \Z^n$. Proof idea: for each $j$, take the dyadic cubes of side $2^{-j}$ that meet the layer $\set{x \in \Omega : 2^{-j+1}\sqrt{n} < \operatorname{dist}(x, F) \le 2^{-j+2}\sqrt{n}}$. These cubes satisfy clause 3 by the triangle inequality, and they cover $\Omega$. Two dyadic cubes are either nested or have disjoint interiors, so keeping only the maximal cubes in the collection gives clause 2.

Clause 3 implies that neighbouring cubes have comparable sizes: if $Q_k$ and $Q_l$ touch, then $\operatorname{diam}(Q_l) \le \operatorname{dist}(Q_l, F) \le \operatorname{dist}(Q_k, F) + \operatorname{diam}(Q_k) \le 5 \operatorname{diam}(Q_k)$. It follows that, for a fixed small enlargement factor such as $1 + \frac{1}{4}$, each point of $\Omega$ lies in at most $C(n)$ of the enlarged cubes.

Example: for $\Omega = (0, \infty) \subseteq \R$ and $F = (-\infty, 0]$, the intervals $[2^{j}, 2^{j+1}]$ for $j \in \Z$ cover $\Omega$ with disjoint interiors. Their distance to $F$ equals their length, so clause 3 holds.

Going deeper

For a student

Whitney introduced this decomposition to extend smooth functions from a closed set $F$ to all of $\R^n$, in the Whitney extension theorem. In harmonic analysis it is applied to the open set $\set{Mf > \lambda}$ in the Calderón–Zygmund decomposition, in the atomic decomposition of Hardy spaces, and in proofs of good-$\lambda$ inequalities. A partition of unity subordinate to the slightly enlarged cubes has derivatives bounded by $C \operatorname{dist}(x, F)^{-k}$, which is what such arguments need.