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Normed versus topological groups: Dichotomy and duality

Normed versus topological groups: Dichotomy and duality

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92 N. H. Bingham <strong>and</strong> A. J. OstaszewskiWe now see that the preceding theorem is applicable to a Haar measure on a locallycompact group X by reference to the following result. Here bounded means pre-compact(covered by a compact set).Theorem 7.3 (Haar measure density theorem – [Mue]; cf. [Hal-M, p. 268]). Let A be aσ-bounded subset <strong>and</strong> µ a left-invariant Haar measure of a locally compact <strong>topological</strong>group X. Then there exists a sequence U n of bounded measurable neigbourhoods of e Xsuch that m ∗ (A ∩ U n x)/m ∗ (U n x) → 1 for almost all x out of a measurable cover of A.Corollary 7.4. In the setting of Theorem 6.4 with A of positive, totally-finite Haarmeasure, let (A, S A , m A ) be the induced probability subspace of X with m A (T ) = m(S ∩A)/m(A) for T = S ∩ A ∈ S A . Then the density theorem holds in A.We now offer a generalization of a result from [BOst-LBII]; cf. Theorem 6.2.Theorem 7.5 (Second Verification Theorem for weak category convergence). Let X bea locally compact <strong>topological</strong> group with left-invariant Haar measure m. Let V be m-measurable <strong>and</strong> non-null. For any null sequence {z n } → e <strong>and</strong> each k ∈ ω,H k = ⋂ n≥k V \(V · z n) is of m-measure zero, so meagre in the d-topology.That is, the sequence h n (x) := xzn−1(wcc)satisfies the weak category convergence conditionProof. Suppose otherwise. We write V z n for V · z n , etc. Now, for some k, m(H k ) > 0.Write H for H k . Since H ⊆ V, we have, for n ≥ k, that ∅ = H ∩ h −1n (V ) = H ∩ (V z n )<strong>and</strong> so a fortiori ∅ = H ∩ (Hz n ). Let u be a metric density point of H. Thus, for somebounded (Borel) neighbourhood U ν u we haveFix ν <strong>and</strong> putm[H ∩ U ν u] > 3 4 m[U νu].δ = m[U ν u].Let E = H ∩ U ν u. For any z n , we have m[(Ez n ) ∩ U ν uz n ] = m[E] > 3 4δ. By TheoremA of [Hal-M, p. 266], for all large enough n, we havem(U ν u△U ν uz n ) < δ/4.Hence, for all n large enough we have m[(Ez n )\U ν u] ≤ δ/4. Put F = (Ez n ) ∩ U ν u; thenm[F ] > δ/2.But δ ≥ m[E ∪ F ] = m[E] + m[F ] − m[E ∩ F ] ≥ 3 4 δ + 1 2δ − m[E ∩ F ]. Som[H ∩ (Hz n )] ≥ m[E ∩ F ] ≥ 1 4 δ,contradicting ∅ = H ∩ (Hz n ). This establishes the claim.

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