16.7. H p AND mp 447

Let Fδ = ∪∞j=1C j. Thus Fδ ⊇ E and

H sδ(E)≤H s

δ(Fδ )≤

∑j=1

β (s)(r(C j))s =

∑j=1

β (s)(r (C j))s < δ +H s

δ(E).

Let δ k→ 0 and let F = ∩∞k=1Fδ k

. Then F ⊇ E and

H sδ k(E)≤H s

δ k(F)≤H s

δ k(Fδ )≤ δ k +H s

δ k(E).

Letting k→ ∞, H s(E)≤H s(F)≤H s(E)We can also arrange to have F containing E be a Gδ set. In 16.10, replace C j with

C j +B(0,η j

)which is an open set having diameter no more than diam(C j)+ 2η j so by

taking η j small enough, we can replace each C j with an open set O j in such a way as topreserve 16.10 with C j replaced with O j and also r (O j)< δ . Then letting Vδ ≡ ∪ jO j,

H sδ(E)≤H s

δ(Vδ )≤

∑j=1

β (s)(r (O j))s < δ +H s

δ(E).

Then let G = ∩kVδ kwhere δ k → 0 and let the Vδ k

be decreasing as k increases, eachVδ containing E. Then for each δ , H s

δ k(E) ≤H s

δ k(G) < δ +H s

δ k(E) Let k→ ∞ to find

H s(E)≤H s(G)≤H s(E) as before. ■A measure satisfying the conclusion of Theorem 16.6.1 is called a Borel regular mea-

sure.

16.7 H p and mp

Next I will compare H p and mp. First recall this covering theorem which is a summary ofCorollary 9.12.5 found on Page 265.

Theorem 16.7.1 Let E ⊆ Rp and let F be a collection of balls of bounded radiisuch that F covers E in the sense of Vitali. Then there exists a countable collection ofdisjoint balls from F , {B j}∞

j=1, such that mp(E \∪∞j=1B j) = 0.

Recall the following interesting lemma stated here for convenience. It is Lemma 11.7.2.

Lemma 16.7.2 Every open set U in Rp is a countable disjoint union of half open boxesof the form Q≡∏

pi=1[ai,ai +2−k) where ai = l2−k for l some integer.

Lemma 16.7.3 If S⊆Rp and mp (S) = 0, then H p (S) =H pδ(S) = 0. Also, there exists

a constant k such that H p (E) ≤ kmp (E) for all E Borel k ≡ β (p)α(p) . Also, if Q0 ≡ [0,1)p,

the unit cube, then ∞ > H p ([0,1)p)> 0.

Proof: Suppose first mp (S) = 0. Without loss of generality, S is bounded. Then byouter regularity, there exists a bounded open V containing S and mp (V ) < ε . For each

x ∈ S, there exists a ball Bx such that B̂x ⊆ V and δ > r(

B̂x

). By the Vitali covering

theorem there is a sequence of disjoint balls {Bk} such that{

B̂k

}covers S. Here B̂k has

16.7. #2? ANDm, 447Let Fs = U3_,Cj. Thus Fs > E andLet 6, + 0 and let F = Ne ¥5,- Then F > E andH5 (E) < H5 (PF) < HG (Fs) < 8 + HG (E)-Letting k + 0, H°(E) < H*(F) < H*(E)We can also arrange to have F containing E be a Gs set. In 16.10, replace C; withC;+B (0, nj) which is an open set having diameter no more than diam (Cj) + 2n ; so bytaking 7 ; small enough, we can replace each C; with an open set O; in such a way as topreserve 16.10 with C; replaced with O; and also r(Oj;) < 6. Then letting Vs = UjO;,MsHB (E) < H3(V5) < Y B(s)(r (Os) < 5+ AG (E).j=lThen let G=1,Vs5, where 6, — 0 and let the Vs, be decreasing as k increases, eachVs containing E. Then for each 6, Hs, (E) < H5 (G) <6 + H5. (E) Let k — to findKOE) < H(G) < H*(E) as before.A measure satisfying the conclusion of Theorem 16.6.1 is called a Borel regular mea-sure.16.7 #? andm,yNext I will compare #? and mp. First recall this covering theorem which is a summary ofCorollary 9.12.5 found on Page 265.Theorem 16.7.1 Let £ C R? and let ¥ be a collection of balls of bounded radiisuch that ¥ covers E in the sense of Vitali. Then there exists a countable collection ofdisjoint balls from F, {Bj}7_\, such that mp(E \ U?_,B;) =0.Recall the following interesting lemma stated here for convenience. It is Lemma 11.7.2.Lemma 16.7.2 Every open set U in R? is a countable disjoint union of half open boxesof the form Q = []f_, [ai,a; +2") where a; = 12~* for | some integer.Lemma 16.7.3 IfS CR? and mp (S) =0, then #? (S) = Hf (S) =0. Also, there existsa constant k such that #0? (E) < kmp (E) for all E Borel k= et. Also, if Qo = [0,1)?,the unit cube, then © > #7? ((0,1)?) > 0.Proof: Suppose first m,(S) = 0. Without loss of generality, S is bounded. Then byouter regularity, there exists a bounded open V containing S and m,(V) < €. For eachax € S, there exists a ball B, such that Bu CVanddé>r (B:). By the Vitali coveringtheorem there is a sequence of disjoint balls {B,} such that {Bi} covers S. Here By has