Kenneth Kuttler

8.7. HARD TOPOLOGY THEOREMS 201

set be contained in the open set(∪n1

j=1B j

)C. It follows there exist disjoint closed balls from

C called Bn1+1, · · · ,Bn2 such that

m̄((

F \∪n1j=1B j

)\∪n2

j=n1+1B j

)< θ pm̄

(F \∪n1

j=1B j

)< θ

2pm̄(F)

continuing this way and noting that limn→∞ θnp = 0 while m̄(F) < ∞, this shows the de-

sired result. Either the process stops because m̄(

F \∪nkj=1B j

)= 0 or else you obtain

m̄(

F \∪∞j=1B j

)= 0. ■

The conclusion holds for arbitrary balls, open or closed or neither. This follows fromobserving that the measure of the boundary of a ball is 0. Indeed, let

S (x,r)≡ {y : |y−x|= r} .

Then for each ε < r,

mp (S (x,r)) ⊆ mp (B(x,r+ ε))−mp (B(x,r− ε))

= mp (B(0,r+ ε))−mp (B(0,r− ε))

((r+ ε

−(

r− ε

)p)(mp (B(0,r)))

Hence mp (S (x,r)) = 0.Thus you can simply omit the boundaries or part of the boundary of the closed balls

and there is no change in the conclusion. Just first apply the above corollary to the Vitalicover consisting of closures of the balls before omitting part or all of the boundaries. Thefollowing theorem is also obtained. You don’t need to assume the set is bounded.

Theorem 8.6.6 Let E be a bounded set and let C be a Vitali covering of E consistingof balls, open, closed, or neither. Let r (B) denote the radius of one of these balls. Thenassume also that sup{r (B) : B ∈ C } = M < ∞. Then there is a countable subset of Cdenoted by {Bi} such that m̄p

(E \∪N

i=1Bi)= 0,N ≤ ∞, and Bi ∩B j = /0 whenever i ̸= j.

Here m̄p denotes the outer measure determined by mp. The same conclusion follows if youomit the assumption that E is bounded.

Proof: It remains to consider the last claim. Consider the balls

B(0,1) ,B(0,2) ,B(0,3) , · · · .

If E is some set, let Er denote that part of E which is between B(0,r−1) and B(0,r) butnot on the boundary of either of these balls, where B(0,−1)≡ /0. Then ∪∞

r=0Er differs fromE by a set of measure zero and so you can apply the first part of the theorem to each Erkeeping all balls between B(0,r−1) and B(0,r) allowing for no intersection with any ofthe boundaries. Then the union of the disjoint balls associated with Er gives the desiredcover. ■

8.7 Hard Topology Theorems8.7.1 The Brouwer Fixed Point TheoremI found this proof of the Brouwer fixed point theorem in Evans [17] and Dunford andSchwartz [15]. The main idea which makes proofs like this work is Lemma 4.7.2 which isstated next for convenience.

8.7. HARD TOPOLOGY THEOREMS 201Ccset be contained in the open set (uit \B i) . It follows there exist disjoint closed balls from@ called By,+41,°+: ,Bny such thatma ((F\UjL Bj) \Uj2n, Bi) < 8pm (F\UFLB;) < 6pm (F)continuing this way and noting that lim, ,.. 67, = 0 while m(F’) < o, this shows the de-sired result. Either the process stops because m (F \U%‘\B i) = 0 or else you obtainm(F\UF_,B)) =0.The conclusion holds for arbitrary balls, open or closed or neither. This follows fromobserving that the measure of the boundary of a ball is 0. Indeed, letS(x,r) = {y:ly—x| =r}.Then for each € <7,mp (S(x,r)) CS mp (B(x,r+€)) — mp (B(x,r—€))= m,(B(0,r+e)) —m,(B(0,r—e))= ((4£) -(F*) )im@onyHence my (S(x,r)) =0.Thus you can simply omit the boundaries or part of the boundary of the closed ballsand there is no change in the conclusion. Just first apply the above corollary to the Vitalicover consisting of closures of the balls before omitting part or all of the boundaries. Thefollowing theorem is also obtained. You don’t need to assume the set is bounded.Theorem 8.6.6 Let E be abounded set and let @ be a Vitali covering of E consistingof balls, open, closed, or neither. Let r(B) denote the radius of one of these balls. Thenassume also that sup{r(B):BE@}=M <o. Then there is a countable subset of ©denoted by {Bj} such that itp (E\ UN_,B;) =0,N < ©, and B} 1B; = 0 whenever i F j.Here mp denotes the outer measure determined by mp. The same conclusion follows if youomit the assumption that E is bounded.Proof: It remains to consider the last claim. Consider the ballsB(0,1),B(0,2),B(0,3),---.If E is some set, let E, denote that part of E which is between B(0,r—1) and B(0,r) butnot on the boundary of either of these balls, where B(0,—1) = 0. Then U™_p£, differs fromE by a set of measure zero and so you can apply the first part of the theorem to each E,keeping all balls between B(0,r—1) and B(0,r) allowing for no intersection with any ofthe boundaries. Then the union of the disjoint balls associated with E, gives the desiredcover. i8.7. Hard Topology Theorems8.7.1 The Brouwer Fixed Point TheoremI found this proof of the Brouwer fixed point theorem in Evans [17] and Dunford andSchwartz [15]. The main idea which makes proofs like this work is Lemma 4.7.2 which isstated next for convenience.