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140 (1991)

A new proof of Kelley’s Theorem

by

Siu-Ah N g (Hull)

Abstract. Kelley’s Theorem is a purely combinatorial characterization of measure algebras. We first apply linear programming to exhibit the duality between measures and this characterization for finite algebras. Then we give a new proof of the Theorem using methods from nonstandard analysis.

First some notation and definitions. B always denotes a Boolean algebra of the form (B, 0, 1, +, ·, −) with the induced ordering ≤. We sometimes write P and Q for + and ·, especially when the operations are infinitary.

By a measure µ on a subalgebra A ⊆ B we mean a finitely additive monotone function µ : A → [0, 1] so that µ(0) = 0 and µ(1) = 1. A Boolean algebra B is called a measure algebra if there is a measure µ : B → [0, 1] which is strictly positive, i.e. µ(b) = 0 iff b = 0. A σ-algebra B is called a σ-measure algebra if there is such a σ-additive µ. Following [K], we define for A ⊆ B the intersection number of A as

α(A) = inf{ α(a b 1 , . . . , a n ) : a 1 , . . . , a n ∈ A, n < ω} , where α(a b 1 , . . . , a n ) = n −1 max{|I| : I ⊆ {1, . . . , n}, Q

i∈I a i 6= 0}. (The a i ’s are not necessarily distinct.) We also define the measure number of A as

β(A) = sup{r ∈ [0, 1] : there is a measure µ on the subalgebra generated by A such that µ(a) ≥ r for all a ∈ A} . A σ-algebra B is said to be weakly ω-distributive if given {b ij : i, j < ω} such that b i,j ≥ b ij+1 , then

ω

X

i=0 ω

Y

j=0

b ij =

ω

Y

n=0 ω

X

i=0

b if

n

(i)

for some f n : ω → ω such that f n ≤ f n+1 (i.e. ∀i, f n (i) ≤ f n+1 (i)).

1991 Mathematics Subject Classification: Primary 03H05, 28A60.

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We assume some basic knowledge of nonstandard analysis (cf. e.g. [HL]

and [L]) and work in a ℵ 1 -saturated nonstandard universe that has the en- largement property (i.e. every standard set has a hyperfinite extension). We use nonstandard analysis and the duality theorem for linear programming to prove the following theorem of J. L. Kelley. See [K] or [F] for a standard proof.

Kelley’s Theorem. (1) A Boolean algebra B is a measure algebra iff (∗) there are A n ⊂ B such that α(A n ) > 0 and B = S

n<ω A n ∪ {0} . (2) A σ-algebra B is a σ-measure algebra iff B is weakly ω-distributive and also satisfies (∗).

The proof is based on the following lemma, which exhibits the duality between measures and Kelley’s characterization.

Lemma. Let A ⊆ B such that A is finite. Then α(A) = β(A) and both the infimum and supremum in the definitions of α and β are attained.

P r o o f. Write A = {a 0 , . . . , a n }, and identify the finite Boolean algebra generated by A as a power set algebra P(X) for some X = {p 0 , . . . , p r }.

For i ≤ r, j ≤ n, define

m ij =  1 if p i ∈ a j , 0 otherwise,

and write M = [m ij ], an (r + 1) × (n + 1) binary matrix.

Claim 1. (a) α(A) is the minimal α ∈ R such that M ·

 x 0

.. . x n

 ≤

 α .. . α

 for some x 0 + . . . + x n ≥ 1, x 0 , . . . , x n ≥ 0.

(b) α(A) = α(a b i

0

, . . . , a i

l

) for some a i

0

, . . . , a i

l

∈ A.

P r o o f. Let (α 0 , . . . , α n , α) ∈ [0, ∞) n+2 be an extreme point in a region determined by some of the r + 2 hyperplanes

m i0 x 0 + . . . + m in x n = x n+1 , i = 0, . . . , r , x 0 + . . . + x n = 1 ,

with the property that α is minimal.

Then this is the minimal α such that M · [x 0 , . . . , x n ] T ≤ α among all x 0 , . . . , x n ≥ 0 satisfying x 0 + . . . + x n ≥ 1. Since the coefficient of each one of the x i is either 0 or 1, the extreme point is a rational point of the form

(α 0 , . . . , α n , α) = (s 0 /s, . . . , s n /s, α) ,

where s 0 , . . . , s n ∈ N and s 0 + . . . + s n = s.

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Any such rational point (s 0 /s, . . . , s n /s) is in one-to-one correspondence with the following sequence from A:

σ = (a 0 . . . a 0 a 1 . . . a 1 . . . a n . . . a n ) , where a 0 repeats s 0 times, a 1 repeats s 1 times, etc.

Moreover, for the above extreme point, we have α(σ) = α, and for any b other sequence τ from A, we have α(τ ) ≥ α. Therefore α(A) = α and (a) b is proved. Since α(A) = α(σ), (b) holds as well. b

Claim 2. (a) β(A) is the maximal β ∈ R such that

M T ·

 y 0

.. . y r

 ≥

 β

.. . β

 for some y 0 + . . . + y r ≤ 1, y 0 , . . . , y r ≥ 0.

(b) β(A) is attained by some measure on the subalgebra generated by A.

P r o o f. Similar to Claim 1, the maximal β satisfying the matrix in- equality is given by some rational point (β 0 , . . . , β r , β). This corresponds to the measure that assigns weights β 0 , . . . , β r to p 0 , . . . , p r respectively. Con- versely, any such weights β 0 , . . . , β r satisfy the matrix inequality for some β, thus the claim is proved.

The linear programming problems in the above two claims are dual to each other, so α(A) = β(A) and the lemma is proved.

Corollary. α(A) ≥ β(A) for arbitrary A ⊆ B.

P r o o f. α(A) = inf α(A 0 ) = inf β(A 0 ) ≥ β(A), where the infimum is taken over all finite A 0 ⊆ A.

P r o o f o f K e l l e y ’ s T h e o r e m (1). (⇒). Let µ : B → [0, 1] be a strictly positive measure and let A n = {b ∈ B : µ(b) ≥ 2 −n }. Then B = S

n<ω A n ∪ {0}. Since β(A n ) ≥ 2 −n , so by the Corollary, α(A n ) > 0.

(⇐). Suppose B = S

n<ω A n ∪ {0} and α(A n ) > 0. Use the enlargement

property and let B 0 be a hyperfinite algebra such that B ⊆ B 0 B. Let

B n = B 0 A n . It follows from { a : a ∈ A n } ⊆ B n A n and the transfer

principle that α(B n ) ≈ α(A n ) > 0 and both are noninfinitesimal. By

transferring the Lemma, α(B n ) = β(B n ), and hence there is an internal

measure ν n on the subalgebra generated by B n such that for each b ∈ B n ,

ν n (b) ≥ α(B n ) > 0, noninfinitesimal. For each m < ω, there is an internal

measure ν on B 0 such that ν ≥ 2 −n ν n on B n for all n < m. So by ℵ 1 -

saturation, there is an internal measure ν on B 0 such that for all n < ω,

ν ≥ 2 −n ν n on B n . Let µ = ν  B, i.e. ∀b ∈ B, µ(b) = ν( b). Then µ is a

strictly positive measure on B.

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P r o o f o f K e l l e y ’ s T h e o r e m (2). (⇒). Let µ : B → [0, 1] be a strictly positive σ-measure. By (1), (∗) holds. To show the weak ω- distributivity, let {b ij } be such that b ij ≥ b i,j+1 . For each n and i < ω, let f n (i) = least j such that

µ

 b ij −

ω

Y

k=0

b ik



≤ 1/2 n+i+1 . Then µ( P ω

i=0 b if

n

(i) − P ω i=0

Q ω

j=0 b ij ) ≤ 1/2 n . Hence

ω

Y

n=0 ω

X

i=0

b if

n

(i) =

ω

X

i=0 ω

Y

j=0

b ij .

(⇐). Let B 0 be a hyperfinite algebra so that B ⊆ B 0 B. Let ν be the internal measure given by the proof in (1). Then ν(b) > 0 for each b ∈ B 0 A n . Let

D = {b ∈ B 0 : there is a decreasing sequence {c n } n<ω from B so that Q ω

n=0 c n = 0 and each c n ≥ b} . Note that Q ω

n=0 c n is defined for the σ-algebra B, while in general, B 0 is not a σ-algebra, and { c n } n<ω may have nonzero lower bounds in B 0 .

Define r = sup{ ν(b) : b ∈ D}.

Claim. r is attained by some a 0 ∈ D. In other words, there are a 0 ∈ B 0 and decreasing d n ∈ B so that Q ω

n=0 d n = 0, each d n ≥ a 0 and r ≈ ν(a 0 ).

P r o o f. Choose b 0 , b 1 , . . . from D so that ν(b i ) → r. For each i, choose c in ∈ B (n < ω) so that c in decreases to 0 as n → ω and each c in ≥ b i . By the weak ω-distributivity, P ω

i=0

Q ω

j=0 c ij = Q ω n=0

P ω

i=0 c if

n

(i) for some f n : ω → ω such that f n ≤ f n+1 . The left hand side equals 0. Write d n = P ω

i=0 c if

n

(i) . Then Q ω

n=0 d n = 0. Note also that d n ∈ B and decreases.

For any n, m, there is a ∈ B 0 such that b 0 + . . . + b m = a ≤ d n , so by ℵ 1 - saturation, there is an internal a 0 ∈ B 0 such that b i ≤ a 0 d n for all i, n < ω (so in particular a 0 ∈ D), and ν(a 0 ) ≥ sup i<ω ν(b i ). Hence ν(a 0 ) = r. The claim is proved.

Notice that ν(a 0 ) 6≈ 1. Suppose otherwise; then for each n < ω, ν( d n )

= 1, so ν(1 − d n ) = 0, and by the strict positivity of ν  B, it follows that d n = 1, contradicting Q ω

n=0 d n = 0.

Now define the internal measure µ(b) = ν(b − a 0 )/(1 − ν(a 0 )), b ∈ B 0 . For b ∈ B, write µ(b) = (µ( b)). Then µ is a strictly positive σ-measure on B. To show this, it suffices to verify the following.

(i) If 0 6= b ∈ B then µ(b) > 0.

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P r o o f. Let d n decrease to 0 in B and d n ≥ a 0 for any n < ω, as in the claim. Since b 6= 0, b is not a lower bound of {d n }, therefore b − d n 6= 0 for some n. Then ν( b − d n ) > 0, so ν( b − a 0 ) > 0, so µ(b) > 0.

(ii) µ is σ-additive on B.

P r o o f. Let {c n } n<ω be a sequence decreasing to 0 in B. By enlarge- ment, this extends to a sequence {c n } n<H in B 0 , where H is an infinite hyperfinite integer, and for convenience, we omit the “ ” from c n for fi- nite n. From the definition of D, c N ∈ D for each infinite N. By a 0 being maximal, for any infinite N, ν(c N + a 0 ) ≈ ν(a 0 ), so ν(c N − a 0 ) ≈ 0. Thus µ(c N ) ≈ 0 for any infinite N. Therefore µ(c n ) → 0.

Acknowledgements. I was supported by an SERC Grant during the writing of this paper. I am very grateful to the unknown referee for his careful reading and very helpful suggestions.

References

[F] D. H. F r e m l i n, Measure algebras, in: Handbook of Boolean Algebra, Vol. III, J. D. Monk and R. Bonnet (eds.), North-Holland, Amsterdam 1989, 877–980.

[HL] A. H u r d and P. A. L o e b, An Introduction to Nonstandard Real Analysis, Aca- demic Press, New York 1985.

[K] J. L. K e l l e y, Measures on Boolean algebras, Pacific J. Math. 9 (1959), 1165–1177.

[L] T. L. L i n d s t r ø m, An invitation to nonstandard analysis, in: Nonstandard Anal- ysis and its Applications, N.J. Cutland (ed.), Cambridge University Press, 1988, 1–105.

DEPARTMENT OF PURE MATHEMATICS UNIVERSITY OF HULL

HULL, HU6 7RX ENGLAND

Received 22 November 1990 ;

in revised form 24 June 1991

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