 By D. H Fremlin

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Example text

Let µY be the subspace measure on Y and ΣY its domain. (a) If (X, Σ, µ) is complete, or totally finite, or σ-finite, or strictly localizable, so is (Y, ΣY , µY ). If Xi i∈I is a decomposition of X for µ, then Xi ∩ Y i∈I is a decomposition of Y for µY . (b) If (X, Σ, µ) has locally determined negligible sets, then µY is semi-finite. (c) If (X, Σ, µ) is complete and locally determined, then (Y, ΣY , µY ) is complete and semi-finite. (d) If (X, Σ, µ) is complete, locally determined and localizable then so is (Y, ΣY , µY ).

Thus E ∈ Σ is an atom for µ with µ ˜(H E) = 0 and µE = µ ˜H < ∞. ˜ and there is an atom E for µ such that µE < ∞ and µ ˜ be a subset (ii) If H ∈ Σ ˜(H E) = 0, let G ∈ Σ of H. We have µ ˜G ≤ µ ˜H = µE < ∞, so there is an F ∈ Σ such that F ⊆ G and µ ˜(G \ F ) = 0. Now either µ ˜G = µ(E ∩ F ) = 0 or µ ˜(H \ G) = ˜ and G ⊆ H; also µ µ(E \ F ) = 0. This is true whenever G ∈ Σ ˜H = µE > 0. So H is an atom for µ ˜. ˜ µ (e) If (X, Σ, µ) is atomless, then (X, Σ, ˜) must be atomless, by (d). ˜ µ If (X, Σ, µ) is purely atomic and H ∈ Σ, ˜H > 0, then there is an E ∈ Σ such that 0 < µ ˆ(H ∩ E) < ∞.

Set E = E0 ∩ E1 ; this serves. 214E Subspaces 37 (b)(i) If A is µY -negligible, there is a set F ∈ ΣY such that A ⊆ F and µY F = 0; now µ∗ A ≤ µ∗ F = 0 so A is µ-negligible, by 132Ad. (ii) If A is µ-negligible, there is an E ∈ Σ such that A ⊆ E and µE = 0; now A ⊆ E ∩ Y ∈ ΣY and µY (E ∩ Y ) = 0, so A is µY -negligible. (c) If A ⊆ X is µ-conegligible, then A ∩ Y is µY -conegligible, because Y \ A = Y ∩ (X \ A) is µnegligible, therefore µY -negligible. If A ⊆ Y is µY -conegligible, then A ∪ (X \ Y ) is µ-conegligible because X \ (A ∪ (X \ Y )) = Y \ A is µY -negligible, therefore µ-negligible. 