Let {Ii λ} be the maximal dyadic cubes such that |F (Iλ i )| >λ. It is enough to prove that, ong>forong> allɛ>0, there is a γ so that, ong>forong> all i and all λ,ν ( {x ∈ I λ i : F ∗ (x) > 2λ, G ∗ (x) ≤ γλ} ) ≤ ɛν(I λ i ). (1.7)Since ν ∈ A ∞ (ω), it is enough to prove (1.7) with the measure ν replaced by ω; and that is whatwe shall do.From this point the proof is essentially identical to the (very short) proof of Theorem A in[Wi] (**on** page 45). Fix Ii λ. We can take G(Iλ i ) ≤ γλ, or there is nothing to prove. Let I be theunique dyadic cube such that Ii λ ∈ N(I). Because of Ii λ ’s maximality, |F (I)| ≤λ. By Lemma 1.8,we may thereong>forong>e take |F (Ii λ )|≤1.1λ, if we choose γ small enough—and of course we do.Now let η>0besosmallthatω({x ∈ Iiλ : d(x, ∂Ii λ) .8λ, G ∗ (x) ≤ γλ} ) ≤ (ɛ/2)ω(I λ i ).But this is just a rescaled versi**on** of Lemma 1.9 (divide h by .8λ). The Theorem 1.1 is proved.QED.2. Lipschitz **domains**.We are going to apply the Theorem 1.1 to a family of functi**on**s {φ (Q) } Q defined **on** a part ofthe boundary of a bounded Lipschitz domain Ω ⊂ R d+1 . That is, ∂Ω is a finite uni**on** of translati**on**sand rotati**on**s of graphs Γ i of functi**on**s ψ i : R d ↦→ R, and there is a fixed M such that each ψ isatisfies |ψ i (x) − ψ i (y)| ≤M|x − y| ong>forong> all x and y.As we said in the introducti**on**, our attenti**on** will be directed at points near ∂Ω, where we canpretend that we are working in a domain lying above a graph Γ i . But first we wish to say a fewwords about the points in Ω that are (relatively) far from the boundary.Recall our definiti**on** of Ω δ ≡{x ∈ Ω: d(x, ∂Ω) ≤ δ}. Given a measure µ defined **on** Ω, writeµ = µ 1 + µ 2 where µ 1 (E) =µ(E \ Ω δ )andµ 2 (E) =µ(E ∩ Ω δ ).We claim that an inequality of the ong>forong>m(∫ ) 1/q (∫1/p|∇u| q dµ 1 ≤ C |f| vdω) p (2.1)Ω∂Ω18

comes almost (but not quite) ong>forong> free.We have, ong>forong> any x/∈ Ω δ ,∫|∇u(x)| ≤C∂Ω|f| dω,where C depends **on** Ω, δ, and our choice of X 0 (which, without loss of generality, we may assumelies in Ω \ Ω δ ). This inequality comes from the fact thatlinked to the inequality:By Hölder’s Inequality,sup |∇u(x)| ≤Cδ −1 sup |u(x)|x/∈Ω δ x/∈Ω δ/2sup |u(x)| ≤C Ω,δ |f| dω.x/∈Ω δ/2∫∂Ω∫∂Ω(∫ ) 1/p (∫|f| dω ≤ |f| p vdω∂Ω∂Ω) 1/p′σdω .Thus, to get (2.1), it is sufficient to haveµ(Ω \ Ω δ ) 1/q (∫∂Ω) 1/p′σdω ≤ c ′ ,where c ′ is small positive c**on**stant depending **on** Ω, δ, andX 0 .A moment’s thought will show that this is just a “global” versi**on** of the sufficient c**on**diti**on**from Theorem 3.1, with ∂Ω (a bounded set, recall) playing the role of a cube Q b ,andΩ\ Ω δpretending to be T (Q b ). The bump functi**on** being integrated against σ is just χ ∂Ω . It has no “tail”because there is no room ong>forong> **on**e:the “cube” Q b fills up all of ∂Ω.To fit the pattern of Theorem 3.1, the c**on**stant c ′ should really be cdiam(Ω)ω(∂Ω) = cdiam(Ω).But, of course, it is—assuming we choose c properly!That is (almost) all we will say about the parts of Ω lying far from the boundary.Our problem now c**on**sists in finding an appropriate c**on**diti**on** **on** the measure µ 2 which ensuresthat(∫(∫) 1/q|∇u| q dµ 2 ≤Ω|f| p vdω∂Ωholds ong>forong> all f ∈ L ∞ , where we remember that µ 2 is supported in Ω δ .By cutting the integral up, taking δ small enough, rescaling, and doing a rotati**on**, we canassume that the support of µ 2 —which we will henceong>forong>th call µ—lies in a set of the ong>forong>m) 1/pR≡{(x, y) : ψ(x)