Helms L.L. Potential Theory

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168 4 Negligible Sets

such that G

ν is bounded and continuous on Ω. By the reciprocity theorem,

Theorem 3.5.1,



μdν =



νdμ<+∞.

Since G

≤ G

,j ≥ 1,



udν = lim

j→∞



dν = lim

j→∞



νdμ



νdμ=



μdν;

that is,



(u −G

μ) dν =0.Sinceu ≥ G

μ on Ω, u = G

μa.e.(ν),u≥ ˆu ≥

μ and therefore u =ˆua.e.(ν). Thus, ν(Γ )=ν(M) = 0. It follows from

Theorem 4.3.9 that Γ is a polar set and that u =ˆu inner quasi everywhere.

Lemma 4.4.11 If Γ ∈K(Ω) and u is a ﬁnite-valued, nonnegative super-

harmonic function on Ω,then

= R

except possibly at points of ∂Γ,

= R

= u q.e. on Γ ,and

= R

q.e. on Ω.

Proof: (Assuming Property B) The ﬁrst assertion follows from Lemma 4.3.2.

By the proved part of Theorem 4.4.10, R

inner quasi everywhere on

Ω.SinceR

= u on Γ,

= u inner quasi everywhere on Γ .Since

Γ ∩{

<u} =

∞

j=1



Γ ∩{

≤ (1 −

)u}



and each term of the union is polar, the countable union is polar by

Theorem 4.2.12.

Lemma 4.4.12 If u is a ﬁnite-valued, positive, superharmonic function on

Ω,then

∪Γ

+ R

∩Γ

≤ R

+ R

(4.1)

and

∪Γ

∩Γ

≤

(4.2)

for all Γ

,Γ

∈K(Ω).

Proof: (Assuming Property B) If x ∈ Γ

∩ Γ

, then (4.1) is true since all

terms are equal to 1. If, for example, x ∈ Γ

∼ Γ

,then

∪Γ

(x)+R

∩Γ

(x)=u(x)+R

∩Γ

(x) ≤ u(x)+R

(x)=R

(x)+R

(x).

By the preceding lemma,

∪Γ

= R

∪Γ

∩Γ

= R

∩Γ

= R

and

= R

quasi everywhere on Γ

∪ Γ

. Thus, (4.2) of the lemma

holds quasi everywhere on Γ

∪ Γ

. By Lemma 3.3.6, the greatest harmonic

minorant of the left side of the latter inequality is the zero function and is

therefore the potential of a measure with support in Γ

∪Γ

. It follows from

the domination principle that

4.4 Capacity 169

∪Γ

∩Γ

≤

on Ω.

Since R

∪Γ

is harmonic on ∼ (Γ

∪ Γ

∪Γ

= R

∪Γ

on ∼ (Γ

∪ Γ

etc. Thus, (4.1) holds on ∼ (Γ

∪ Γ

), and since it already has been seen to

hold on Γ

∪ Γ

, it holds on Ω.

Lemma 4.4.13 If u is a ﬁnite-valued, nonnegative superharmonic function

on Ω and Γ ∈K(Ω),then

is the greatest potential of measures μ with

support in Γ such that G

μ ≤ u.

Proof: (Assuming Property B) Let μ be any measure with compact support

in Γ satisfying G

μ ≤ u on Ω.Sinceu is ﬁnite-valued on Ω, G

μ<+∞

on Ω.Ifv ∈ Φ

,thenG

μ ≤ u ≤ v on Γ so that G

μ ≤ v on Ω by the

domination principle Theorem 4.4.8. Thus, R

≥ G

μ on Ω and since G

is l.s.c. on Ω,

≥ G

μ.Since

is the potential of a measure ν with

support in Γ,

= G

ν ≥ G

μ for all measures μ with support in Γ .

Lemma 4.4.14 Let u be a ﬁnite-valued, continuous, positive superharmonic

function on Ω and let {Γ

} be a monotone sequence in K(Ω) with limit Γ ∈

K(Ω).

(i) If {Γ

} is an increasing sequence, then lim

j→∞

= R

and

lim

j→∞

on Ω.

(ii) If {Γ

} is a decreasing sequence, then lim

j→∞

= R

on Ω.

Proof: (Assuming Property B) Suppose ﬁrst that {Γ

} is an increasing se-

quence. Since R

= u on Γ

and R

= u on Γ, lim

j→∞

= R

on Γ .

Since R

≤ R

on Ω,lim

j→∞

≤ R

on Ω ∼ Γ and lim

j→∞

≤ R

on Ω.Since

= R

q.e. on Ω,

= R

q.e. on Ω, and the union of

countably many polar sets is again polar, lim

j→∞

≤

on Ω.Since

lim

j→∞

is superharmonic on Ω by Theorem 2.4.8,

= lim

j→∞

q.e. on Γ ,and

is a potential,

≤ lim

j→∞

on Ω by the domi-

nation principle, Theorem 4.4.8. Therefore, lim

j→∞

on Ω.Since

= R

and

= R

on Ω ∼ Γ, lim

j→∞

= R

on Ω.Nowlet{Γ

}

be decreasing sequence of compact subsets of Ω. Deﬁning w = lim

j→∞

on Ω and applying Theorem 2.4.9, the lower regularization ˆw of w is su-

perharmonic on Ω. By the Riesz decomposition theorem, ˆw = G

μ + h

where μ is a measure on Ω and h is the greatest harmonic minorant of

ˆw.Since0≤ w ≤ R

on Ω, 0 ≤ ˆw ≤

on Ω.Since

is a po-

tential, the greatest harmonic minorant of ˆw is the zero function; that is,

h =0and ˆw = G

μ.Forﬁxedj

≥ 1, consider the functions R

,j ≥ j

which are harmonic on Ω ∼ Γ

. By Theorem 2.2.5, w is harmonic and

therefore equal to ˆw on Ω ∼ Γ

.Sincej

≥ 1 is arbitrary, ˆw is har-

monic on Ω ∼ Γ . By Theorem 3.4.6, the support of μ is contained in Γ .

170 4 Negligible Sets

Since ˆw = G

μ ≤ u on Ω,ˆw ≤

by the preceding lemma. Since

≥ R

≥

for all j ≥ 1,w ≥

. Taking the lower regularization

of both sides, ˆw ≥

and consequently ˆw =

.Sincew is harmonic on

Ω ∼ Γ, ˆw = w = lim

j→∞

≥ R

≥

and R

= lim

j→∞

Ω ∼ Γ .SinceR

= u on Γ and R

= u on Γ , R

= lim

j→∞

on Ω.

Example 4.4.15 Part (ii) of the preceding lemma is not true for regularized

reduced functions. For example, if u =1andΓ

= {x; |x|≤1/j},then

1 = lim

j→∞

(0) =0=

{0}

(0).

It should be remembered that R



and



are functions of two variables x

and Γ . Although the variable x will be suppressed in the following, it should

be kept in mind that x is ﬁxed. Except for the last item in the following list,

the following properties of R

and

were proved in the preceding section.

(i) 1 ≥ R

≥

≥ 0onΩ.

(ii) R

=1onΓ .

(iii) 1 = R

on the interior of Γ .

(iv) R

on Ω ∼ Γ and both are harmonic on Ω ∼ Γ .

(v)

is the potential of a measure with support in Γ .

The last fact follows from Theorems 3.4.4 and 3.4.9. It has been seen that it

is possible to have

(x) < R

(x)forsomex. It was shown in Lemma 4.4.11

that the set for which this is true is polar, assuming Property B. It will take

several steps to show that the set of such points is a negligible set for Greenian

sets in general.

Deﬁnition 4.4.16 A nonnegative, extended real-valued set function φ de-

ﬁned on a class of sets F is

(i) strongly subadditive on F if F is closed under ﬁnite unions and ﬁnite

intersections and A, B ∈F⇒φ(A ∪ B)+φ(A ∩ B) ≤ φ(A)+φ(B), and

(ii) countably strongly subadditive on F if F is closed under countable

unions and countable intersections and

φ(∪

∞

j=1

∞



j=1

φ(A

) ≤

∞



j=l

φ(B

)+φ(∪

∞

j=1

)

whenever {A

} and {B

} are sequences in F satisfying A

⊂ B

,j ≥ 1.

Lemma 4.4.17 The nonnegative, ﬁnite-valued set function C(Γ ) deﬁned on

K(Ω) has the following properties.

(i) (Monotonicity) C(∅)=0and Γ

⊂ Γ

⇒C(Γ

) ≤C(Γ

(ii) If {Γ

} is a monotone sequence in K(Ω) with limit Γ ∈K(Ω),then

lim

j→∞

C(Γ

)=C(Γ ).

(iii) (Strong subadditivity) C(Γ

∪ Γ

)+C(Γ

∩Γ

) ≤C(Γ

)+C(Γ

4.4 Capacity 171

Proof: (Assuming Property B) Only (ii) will be proved. The same method

can be used to prove (i) and (iii) using Theorem 4.3.5 and Lemma 4.4.12,

respectively. Note that R

→ R

by the preceding lemma. Let Λ be an open

set having compact closure Λ

−

with ∪Γ

⊂ Λ ⊂ Λ

−

⊂ Ω,andlet

−

= G

where ν is a measure with compact support in ∂Λ.Thus,

= R

and

= R

a.e.(ν) since these functions diﬀer only at points in Λ.Bythe

reciprocity theorem, Theorem 3.5.1, C(Γ



1 dμ



νdμ



dν =



dν =



dν →



dν. Reversing the steps, the

latter integral is just C(Γ )andsoC(Γ

) →C(Γ )asj →∞.

It is apparent that C

∗

(E) is an increasing function of E ⊂ Ω and that

∗

(Γ )=C(Γ ) for all compact sets Γ ⊂ Ω. At the beginning of this section, a

subset E of the Greenian set Ω was deﬁned to be capacitable if C

∗

(E)=C

∗

(E)

and the capacity C(E) was deﬁned to be the common value. It is easy to see

that C

∗

(E) ≤C

∗

(E) for all E ⊂ Ω and that the open sets are capacitable.

Remark 4.4.18 The part of (ii) of the preceding lemma pertaining to mono-

tone decreasing sequences is equivalent to the following assertion.

(ii



)(Right-continuity)IfΓ ∈K(Ω) Ω and >0, then there is a neigh-

borhood Λ of Γ such that C(Σ) −C(Γ ) <for all Σ ∈K(Ω) satisfying

Γ ⊂ Σ ⊂ Λ.

To see that (ii) implies (ii



), let {Λ

} be a decreasing sequence of open sets

such that Γ ⊂ Λ

,j ≥ 1, and Γ = ∩Λ

−

.By(ii), given >0thereisa

≥ 1 such that C(Λ

−

) < C(Γ )+. For any compact set Σ with Γ ⊂ Σ ⊂

, C(Σ) ≤C(Λ

−

) < C(Γ )+ and C(Σ) −C(Γ ) <. Assuming (ii



), let

{Γ

} be a decreasing sequence of compact subsets of Ω with Γ = ∩Γ

.Given

>0, let Λ be a neighborhood of Γ such that C(Σ) −C(Γ ) <whenever

Σ is a compact set with Γ ⊂ Σ ⊂ Λ.Thereisthenaj

≥ 1 such that

Γ ⊂ Γ

⊂ Λ for all j ≥ j

so that C(Γ ) ≤C(Γ

) < C(Γ )+ for all j ≥ j

;

that is, lim

j→∞

C(Γ

)=C(Γ ).

It was noted previously that all open sets are capacitable and that C(Γ )=

∗

(Γ ) for all Γ ∈K(Ω). Note that if O ∈O(Ω)andΓ is a compact subset

of O,then

∗

(O)=sup{C(K); K ∈K(Ω),Γ ⊂ K ⊂ O}.

Lemma 4.4.19 Compact and open subsets of Ω are capacitable.

Proof: (Assuming Property B) By (ii



) of the preceding remark, given >0

thereisanopensetΛ such that C(Σ) −C(Γ ) <for any compact set Σ

satisfying Γ ⊂ Σ ⊂ Λ.Thus,

∗

(Γ ) ≤C

∗

(Λ)=sup{C(Σ); Σ ∈K(Ω),Γ ⊂ Σ ⊂ Λ}≤C(Γ )+.

172 4 Negligible Sets

If O is any open set satisfying Γ ⊂ O ⊂ Λ,thenC

∗

(Γ ) ≤C

∗

(O) ≤C

∗

(Λ)and

C(Γ )=C

∗

(Γ ) ≤ inf {C

∗

(O); O ∈O(Ω),Γ ⊂ O}

=inf{C

∗

(O); O ∈O(Ω),Γ ⊂ O ⊂ Λ}

≤C

∗

(Λ) < C(Γ )+.

But since the ﬁrst inﬁmum is just C

∗

(Γ )and is arbitrary, C(Γ )=C

∗

(Γ )=

∗

(Γ ).

Extending the conclusion of this lemma to Borel sets requires several steps.

Lemma 4.4.20 If A

,...,A

,...,B

∈K(Ω) with A

⊂ B

,j =

1,...,m,then

C(∪

j=1

) −C(∪

j=1

) ≤



j=1

(C(B

) −C(A

)).

Proof: (Assuming Property B) Consider the m = 2 case ﬁrst. Taking Γ

and Γ

= A

∪ B

in (iii) and using (i) of Lemma 4.4.17,

C(B

∪B

)+C(A

) ≤C(B

∪B

)+C(A

∪(B

∩B

)) ≤C(B

)+C(A

∪B

);

also taking Γ

= A

∪ A

and Γ

= B

C(A

∪B

)+C(A

) ≤C(A

∪B

)+C((A

∪B

) ∪A

) ≤C(A

∪A

)+C(B

Thus,

C(B

∪ B

)+C(A

) ≤C(B

)+C(A

∪B

)+C(A

)

≤C(B

)+C(A

∪A

)+C(B

)

and the assertion is true for m = 2. Assume the assertion is true for m − 1.

By the m =2case,

C(∪

j=1

) −C(∪

j=1

)=C(∪

m−1

j=1

∪B

) −C(∪

m−1

j=1

∪ A

)

≤ (C(∪

m−1

j=1

) −C(∪

m−1

j−1

)) + (C(B

) −C(A

))

≤

m−1



j=1

(C(B

) −C(A

)) + (C(B

) −C(A

))



j=1

(C(B

) −C(A

))

and the assertion is true for m whenever it is true for m −1. The conclusion

follows by induction.

Theorem 4.4.21 The outer capacity has the following properties.

(i) A, B ⊂ Ω,A ⊂ B ⇒C

∗

(A) ≤C

∗

(B).

4.4 Capacity 173

(ii) A

⊂ Ω, j ≥ 1,A

↑ A ⇒C

∗

) →C

∗

(A).

(iii) A

∈K(Ω),j ≥ 1,A

↓ A ⇒C

∗

) →C

∗

(A).

(iv) A

⊂ Ω, A

⊂ B

,j ≥ 1 ⇒

∗

(∪

∞

j=1

∞



j=1

∗

) ≤

∞



j=1

∗

)+C

∗

(∪

∞

j=1

Proof: (Assuming Property B) Assertion (i) is trivial and assertion (iii)is

the same as (ii) of Lemma 4.4.17. The other assertions will require several

steps.

Step 1. O

∈O(Ω),j ≥ 1,O

↑ O ⇒C(O

) →C(O).

For each j ≥ 1, let O

= ∪

∞

k=1

with Γ

∈K(Ω), let Σ

= ∪

i,k≤ j

,j ≥ 1,

and let Γ be any compact subset of O.Thus,eachΣ

is compact, Σ

↑ O,

and Σ

∩ Γ ↑ Γ .Since

C(O

)=C



∪

∞

k=1



= C



∪

i≤j

∪

∞

k=1



≥C



∪

i,k≤ j

∩Γ



= C(Σ

∩Γ ),

by (ii) of Lemma 4.4.17

lim

j→∞

C(O

) ≥ lim

j→∞

C(Σ

∩ Γ )=C(Γ ).

Since Γ is any compact subset of O,

lim

j→∞

C(O

) ≥C

∗

(O)=C(O).

As the opposite inequality is always true, lim

j→∞

C(O

)=C(O).

Step 2. A

,...,A

,...,B

⊂ Ω,A

⊂ B

,j =1,...,m⇒

∗

(∪

j=1



j=1

∗

) ≤



j=1

∗

)+C

∗

(∪

j=1

). (4.3)

This inequality will be proved for the m = 2 and open sets case ﬁrst. Let

be open sets with U

⊂ V

,j =1, 2. Let Γ

,Γ

,andΣ be

compact subsets with Γ

⊂ U

,Γ

⊂ U

,andΣ ⊂ V

∪ V

.ThesetsΣ ∼ V

and Σ ∼ V

are disjoint closed sets with Σ ∼ V

⊂ V

and Σ ∼ V

⊂ V

There are disjoint open sets O

and O

such that Σ ∼ V

⊂ O

and Σ ∼

⊂ O

. Letting Σ

= Σ ∼ O

and Σ

= Σ ∼ O

,Σ = Σ

∪ Σ

where Σ

and Σ

are compact subsets of V

and V

, respectively. By Lemma 4.4.20,

C((Σ

∪ Γ

) ∪ (Σ

∪ Γ

)) + C(Γ

)+C(Γ

)

≤C(Σ

∪ Γ

)+C(Σ

∪Γ

)+C(Γ

∪ Γ

174 4 Negligible Sets

Since Σ ⊂ (Σ

∪Γ

) ∪ (Σ

∪ Γ

)andΣ ∼ O

⊂ V

,i=1, 2,

C(Σ)+C(Γ

)+C(Γ

) ≤C(V

)+C(V

)+C(U

∪U

Taking the supremum over Γ

,Γ

,andΣ,

C(V

∪V

)+C(U

) ≤C(V

)+C(V

)+C(U

∪ U

)

so that Inequality (4.3) is true for m = 2 and open sets. As in the proof

of Lemma 4.4.20, Inequality (4.3) is true for all open sets and m ≥ 2.

Consider now arbitrary sets A

,j =1,...m.IfsomeC

∗

)=+∞ or

∗

(∪

j=1

)=+∞, then Inequality (4.3) is trivially true. It therefore can be

assumed that all terms on the right side of the inequality are ﬁnite. If >0,

there is an open set O ⊃∪

j=1

and open sets O

⊃ B

,j =1,...,m,such

that



j=1

C(O

)+C(O) ≤



j=1

∗

)+C

∗

(∪

j=1

)+.

Letting U

= O

∩O ⊂ O

,j =1,...,m,

C(∪

j=1



j=1

C(U

) ≤C(∪

j=1



j=1

C(O

Since ∪

j=1

⊂ O,

C(∪

j=1



j=1

C(U

) ≤C(O)+



j=1

C(O

)

≤C

∗

(∪

j=1



j=1

∗

)+.

Since ∪

j=1

⊂∪

j=1

, the ﬁrst term on the left is greater than or equal

to C

∗

(∪

j=1

); since A

⊂ B

⊂ O

and A

⊂ O, A

⊂ O

∩ O =

,j =1,...,m, and the second term on the left is greater than or equal



j=1

∗

). Therefore,

∗

(∪

j=1



j=1

∗

) ≤C

∗

(∪

j=1



j=1

∗

)+.

Since  is arbitrary, the assertion is true for m ≥ 2 and arbitrary subsets of Ω.

Step 3. A

⊂ Ω, j ≥ 1,A

↑ A ⇒C

∗

) →C

∗

(A).

This result is trivial if lim

j→∞

∗

)=+∞. It therefore can be assumed

that the limit is ﬁnite. If >0, for each j ≥ 1anopensetO

⊃ A

can be

chosen so that C(O

) < C

∗

)+/2

. By the preceding step of the proof

4.4 Capacity 175

C(∪

j=1

) ≤C

∗

(∪

j=1



j=1

(C(O

) −C

∗

))

< C

∗

)+.

By the ﬁrst step of this proof, C(∪

j=1

) →C(∪

∞

j=1

)and

∗

(A)=C

∗

(∪

∞

j=1

) ≤C(∪

∞

j=1

) ≤ lim

m→∞

∗

)+.

Since  is arbitrary and lim

m→∞

∗

) ≤C

∗

(A), C

∗

(A) = lim

m→∞

∗

This completes the proof of (ii).

Step 4. Assertion (iv) follows from Inequality (4.3) and the preceding step

by letting m →∞.

In order to show that the class of capacitable sets includes the Borel sets,

an excursion into set theory is required. After this excursion, Borel sets will

be shown to be capacitable along with certain sets called analytic sets.

Consider a pair (X, F)whereX is some nonempty set and F is a collection

of subsets of X that includes the empty set. F is called a paving of X and

the pair (X, F) is called a paved space.LetN

∞

= N ×N ×··· be the set

of all inﬁnite sequences of positive integers. Elements of N

∞

will be denoted

by n =(n

,...).

Deﬁnition 4.4.22 Let (X, F) be a paved space.

(i) A Suslin scheme is a map (n

,...,n

) → X

...n

∈Ffrom the set of

all ﬁnite sequences of positive integers into F.

(ii) The nucleus of the Suslin scheme is the uncountable union

n∈N

∞

∩ X

∩···);

this set is said to be analytic over F.

(iii) The class of sets analytic over F will be denoted by A(F).

It is easy to see that F⊂A(F) as follows. If A ∈F, for each ﬁnite sequence

,...,n

) of positive integers let X

...n

= A.Themap(n

,...,n

) →

...n

is then a Suslin system and

A =

n∈N

∞

∩ X

∩···)

so that A ∈A(F). To show that a subset B of X belongs to A(F), it must

be shown that there is a Suslin system (n

,...,n

) → X

···n

so that

B =

n∈N

∞

∩ X

∩···).

176 4 Negligible Sets

Lemma 4.4.23 A(F) is closed under countable unions and countable inter-

sections.

Proof: Let {A

} be a sequence in A(F), and let A = ∪A

.Foreachj ≥ 1,

let (n

,...,n

) → X

(j)

···n

be a Suslin scheme with

n∈N

∞

(j)

∩ X

(j)

∩ X

(j)

∩···).

Consider a one-to-one and onto map α : N → N × N.If(m

···m

)isa

ﬁnite sequence of positive integers, deﬁne X

···m

= X

(j)

···n

if α(m

(j, n

),m

= n

,i=2,...k.Thus,

A =

n∈N

∞

∩ X

∩X

∩···)

and A ∈A(F). Therefore, A(F) is closed under countable unions. Now

let {B

} be sequence in A(F), and let B = ∩B

.Foreachj ≥ 1, let

,...,n

) → X

(j)

···n

be a Suslin system with

∈N

∞

(j)

∩ X

(j)

∩ X

(j)

∩···)

where n

=(n

,...). Consider any x ∈∩B

.Toeachsuchx there

corresponds a sequence of sequences {n

} such that

x ∈

j≥1

(j)

∩ X

(j)

∩ X

(j)

∩···).

Arrange the sequence of sequences {n

} into a single sequence m = { m

} by

applying a diagonal procedure so that

m =(n

,...).

Given the latter sequence, the sequence of sequences {n

} can be recon-

structed. Now let

= X

(1)

= X

(2)

= X

(1)

= X

(3)

,....

Then

B ⊂

m∈N

∞

∩ X

∩···).

Conversely, any x in the latter set can be seen to belong to B by constructing

a sequence of sequences {n

} from the {m

} sequence as indicated above.

Thus, B ∈A(F) and the latter is closed under countable intersections.

4.4 Capacity 177

If X is a metric space and F is the collection of closed subsets of X,

then A(F) contains the open sets as well as the closed sets. If, in addition,

X is σ-compact and F is the collection of compact subsets of X,thenA(F)

contains all open sets. Although closed under countable unions and countable

intersections in both cases, A(F) is not necessarily a σ-algebra since it may

not be closed under complementation.

Theorem 4.4.24 If X is a σ-compact metric space and F is the class of

compact subsets of X,thenA(F) contains all Borel subsets of X.

Proof: Let A

(F) be the collection of all ﬁnite unions of ﬁnite intersections

of elements of F and complements of such sets. Then A

(F)isanalgebra

and F⊂A

(F) ⊂A(F). Since A(F) is a monotone class (that is, closed

under countable unions and countable intersections), by a standard theorem

of measure theory it must contain the smallest σ-algebra containing A

(F)

(c.f., Halmos [27] ), since the latter contains all compact sets, A(F) ⊃ σ(F)=

class of Borel subsets of X.

Lemma 4.4.25 (Sierpinski) Let (X, F) be a paved space, let A be the nu-

cleus of a Suslin scheme (n

,...,n

) → X

···n

∈F,andlet{b

} be a

sequence in N. Deﬁne

{n:n

≤b

,i≤j}

∩X

∩···)

{n:n

≤b

,i≤j}

∩···∩X

···n

)

B = ∩

j≥1

Then A

⊂ A, A

⊂ B

j+1

⊂ B

,j ≥ 1,andB ⊂ A.

Proof: All of the conclusions are easy to verify except for the last, namely

that B ⊂ A.Consideranyx ∈ B.Foreachk ≥ 1, there is a k-tuple

,...n

)withn

≤ b

,i≤ k such that

x ∈ X

∩ X

∩···∩X

···n

Consider a ﬁxed sequence of such k-tuples corresponding to x.Sincen

∈

{1, 2,...b

},k ≥ 1, there is an integer m

∈{1,...,b

} such that n

= m

for inﬁnitely many k and x ∈ X

= X

for all such k. Considering only

those k-tuples having m

as the ﬁrst integer, there is an m

∈{1,...,b

}

such that m

is the second integer for inﬁnitely many k-tuples having m

the ﬁrst integer so that x ∈ X

∩ X

= X

∩ X

for all such k.

By an induction argument, there is a sequence {m

},withm

≤ b

,j ≥ 1,

such that

x ∈ X

∩ X

∩··· ⊂A.