# Measure theory/ 3, Measure algebras by D.H. Fremlin

By D.H. Fremlin

Fremlin D.H. degree concept, vol.3 (2002)(ISBN 0953812936)(672s)-o

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Similarly, H ⊆ H2 , and f −1 [H] ⊆ f −1 [H1 ∩ H2 ] ⊆ π(H1 ∩ H2 ). But also H ∩ f [G] is not empty, so ∅ = G ∩ f −1 [H] ⊆ G ∩ π(H1 ∩ H2 ), which is impossible. X XQ Q (c) If H ∈ RO(Y ) and H = Y \ H is its complement in RO(Y ) then πH = X \ πH is the complement of πH in RO(X). P P By (b), πH and πH are disjoint. Now H ∪ H is a dense open subset of Y , so πH ∪ πH ⊇ f −1 [H] ∪ f −1 [H ] = f −1 [H ∪ H ] is dense in X, and the regular open set πH must include the complement of πH in RO(X). Q Q Putting (b) and (c) together, we see that the conditions of 312H(ii) are satisfied, so that π is a Boolean homomorphism.

Then Σ ∩ I is a σ-ideal of the Boolean algebra Σ, and Σ/Σ ∩ I is Dedekind σ-complete. proof Of course Σ is Dedekind σ-complete, because if En n∈N is any sequence in Σ then n∈N En is the least upper bound of {En : n ∈ N} in Σ. It is also easy to see that Σ ∩ I is a σ-ideal of Σ, since F ∩ n∈N En ∈ I whenever F ∈ Σ and En n∈N is a sequence in Σ ∩ I. So 314C gives the result. 314E Proposition Let A be a Boolean algebra. (a) If A is Dedekind complete, then all its order-closed subalgebras and principal ideals are Dedekind complete.

Q Q 34 Boolean algebras 313G (iii) Consequently c \ b ∈ B whenever b, c ∈ B and b ⊆ c. P P If a ∈ B, then b ∩ a, c ∩ a ∈ B ⊆ A and b ∩ a ⊆ c ∩ a, so (c \ b) ∩ a = (c ∩ a) \ (b ∩ a) ∈ A by the hypothesis on A. By (ii), c \ b ∈ B. Q Q (iv) It follows that B is a subalgebra of A. P P If b ∈ B, then b ⊆ 1 ∈ I ⊆ B, so 1 \ b ∈ B. If a, b ∈ B, then a ∪ b = 1 \ ((1 \ a) ∩ (1 \ b)) ∈ B. 0 = 1 \ 1 ∈ B, so that the conditions of 312B(ii) are satisfied. Q Q Now the subalgebra of A generated by I is included in B and therefore in A, as required.