The first difference is ≤ CG(I) ≤ G(I ∗ ), by Lemmas 1.5 and 1.6. The sec**on**d is less than or equalto∑|λ Q ||φ (Q) (x I ∗)|;Q:Q⊂IQ⊄I ∗which, by the Cauchy-Schwartz inequality (see the proof of Lemma 1.5) is less than or equal to⎛G(I ∗ ⎜∑) ⎝Q:Q⊂IQ⊄I ∗⎡⎤⎞∞∑ω(Q) ⎣2 −jτω(2 j Q) χ R j (Q)(x I ∗) ⎦⎟⎠j=0But the first part of the proof of Lemma 1.5 shows that the sec**on**d factor is bounded by a c**on**stant.QED.We are now ready to prove Lemma 1.9, from which Theorem 1.1 will follow as a corollary.Lemma 1.9. Let Q 0 ∈Dbe the unit dyadic cube, and let {φ (Q) } Q be a familyof functi**on**ssatisfying 1), 2), and 3). Let f = ∑ Q λ Qφ (Q) be a finite linear combinati**on** such that λ Q =0ong>forong>all Q not c**on**tained in Q 0 . Then: For all δ>0, there is a γ>0 such thatω ({x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}) ≤ δω(Q 0 ). (1.4)Remark .Ifν ∈ A ∞ (ω), then Lemma 1.9 immediately implies, as a corollary, the same c**on**clusi**on**ong>forong> ν; i.e., ong>forong> all δ>0 there is a γ>0 such thatν ({x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}) ≤ δν(Q 0 ).We will use this corollary to obtain Theorem 1.1.Proof of Lemma 1.9. Let A>1 be a large number, to be chosen presently. Let {I k } k bethe family of maximal dyadic subcubes of Q 0 having the property that, ong>forong> some I ∗ ∈ N(I k ),G(I ∗ ) >Aγ.By Lemmas 1.6 and 1.7, if A is chosen large enough, the set {x ∈ Q 0 : G ∗ (x) >γ} will c**on**tain∪ k I k . We henceong>forong>th assume that A has been chosen ‘large enough.’ On the other hand, noticethat, if x/∈∪ k I k , then G ∗ (x) ≤ Aγ:this will be important.With {I k } k now fixed, let {J l } l be the maximal subcubes of Q 0 such that, first, no J l isc**on**tainedinanyI k , and sec**on**d, |F (J l )| > 1. Denote the uni**on** {I k }∪{J l } by P, andlet{P i } i bethe family of maximal cubes from P.We claim that{x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ} ⊂∪ i P i . (1.5)To see this, suppose that x bel**on**gs to the left-hand side of (1.5). Since F ∗ (x) > 1, x must bel**on**gto some cube J such that |F (J)| > 1. If this cube J were c**on**tained in some I k , then we would haveG ∗ (x) >γ, a c**on**tradicti**on**. Thereong>forong>e, x bel**on**gs to **on**e of the special cubes J l .But∪ l J l ⊂∪ i P i .121/2.

Thus, our problem has now reduced to c**on**trolling the size of ∑ i: |F (P i )|>1 ω(P i).The reader may w**on**der why we throw the cubes I k into P at all, since **on**ly the cubes J l areneeded to cover {x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}. The reas**on** will so**on** become apparent. But,essentially:we use the family {I k } k to c**on**trol the size of G ∗ (x) globally **on** Q 0 (i.e., even at pointswhere F ∗ (x) ≤ 1). This is very much in the spirit of the proof of the classical good-λ inequalityong>forong> the dyadic square functi**on**.Define F 1 = {Q ⊂ Q 0 : ∀i(Q ⊄ P i )} and F 2 = {Q ⊂ Q 0 : Q ⊂ P i ong>forong> some i}; andsetf i = ∑ Q∈F iλ Q φ (Q) ong>forong> i =1, 2. It is obvious that f = f 1 + f 2 . Corresp**on**ding to f 1 and f 2 ,wedefineF i (I,x)=∑λ Q φ (Q) (x)Q:Q∈S(I)Q∈F iF i (I) =F i (I,x I )Fi ∗ (x) = sup |F i (I)|I:x∈I⎛⎡⎜∑∞∑G i (I,x)= ⎝ |λ Q | 2 ⎣Q∈S(I)Q∈F ij=02 −j(2α−τ)⎤⎞ω(2 j Q) χ R j (Q)(x) ⎦⎟⎠G i (I) =G i (I,x I )G ∗ i (x) = sup G i (I);I:x∈Iwhere, as beong>forong>e, we do not define F i (I,x)orG i (I,x)ong>forong>x/∈ I.Beong>forong>e going **on**, let us note—what is easy to see—that F (I,x) =F 1 (I,x)+F 2 (I,x) andG(I,x) ≤ G 1 (I,x)+G 2 (I,x). It is also easy to see that that each G i (I,x) ≤ G(I,x).For any cube Q, defineC(Q) ≡{x ∈ Q : |x − x Q | 1ω(C(P i )).1/2Clearly,∑∑∑ω(C(P i )) ≤ω(C(P i )) +ω(C(P i ))i: |F (P i )|>1i: |F 1 (P i )|>1/2i: |F 2 (P i )|>1/2≡ (I)+(II).Let us c**on**sider (I) first. EachP i satisfies G(P i ) ≤ γ. Thereong>forong>e, if x ∈ C(P i ), we have (byLemma 1.5) |F 1 (P i ) − F 1 (P i ,x)| ≤Cγ. If we take γ small enough, then this difference will be lessthan 1/4, and having |F 1 (P i )| > 1/2 will ong>forong>ce |F 1 (P i ,x)| > 1/4 **on**allofC(P i ). Let us assumethat γ is so chosen. We get:∑ω(C(P i )) ≤ ∑ ω({x ∈ C(P i ): |F 1 (P i ,x)| > 1/4}).i: |F 1 (P i )|>1/2 i13