Normed versus topological groups: Dichotomy and duality

Normed groups 119groups need not be paracompact, as exemplified again by the example due to Oleg Pavlov[Pav] quoted earlier or by the example of van Douwen [vD] (see also [Com, Section 9.4p. 1222] ); however, L. G. Brown [Br-2] shows that a locally complete topological groupis paracompact (and this includes the locally compact case, cf. [Com, Th. 2.9 p. 1161]).The assumption of paracompactness is thus natural.Theorem 11.14 (The Second Generalized Kestelman-Borwein-Ditor Theorem: MeasurableCase – cf. Th. 7.6). Let G be a paracompact topological group equipped with a locallyfinite, inner regular Borel measure m (Radon measure) which is left-invariant, resp. rightinvariant(for example, G locally compact, equipped with a Haar measure).If A is a (Borel) measurable set with 0 < m(A) < ∞ and z n → e, then, for m-almostall a ∈ A, there is an infinite set M a such that the corresponding right-translates, resp.left-translates, of z n are in A, i.e., in the first case{z n a : n ∈ M a } ⊆ A.Proof. Without loss of generality we conside right-translation of the sequence {z n }. SinceG is paracompact, it suffices to prove the result for A open and of finite measure. Byinner-regularity A may be replaced by a σ-compact subset of equal measure. It thussuffices to prove the theorem for K compact with m(K) > 0 and K ⊆ A. Define adecreasing sequence of compact sets T k := ⋃ n≥k z−1 n K, and let T = ⋂ k T k. Thus x ∈ Tiff, for some infinite M x ,z n x ∈ K for m ∈ M x ,so that T is the set of ‘translators’ x for the sequence {z n }. Since K is closed, for x ∈ T,we have x = lim n∈Mx z n x ∈ K; thus T ⊆ K. Hence, for each k,m(T k ) ≥ m(z −1kK) = m(K),by left-invariance of the measure. But, for some n, T n ⊆ A. (If zn−1 k n /∈ A on an infiniteset M of n, then since k n → k ∈ K we have zn−1 k n → k ∈ A, but k = lim zn −1 k n /∈ A,a contradiction since A is open.) So, for some n, m(T n ) < ∞, and thus m(T k ) → m(T ).Hence m(K) ≥ m(T ) ≥ m(K). So m(K) = m(T ) and thus almost all points of K aretranslators.Remark. It is quite consistent to have the measure left-invariant and the metric rightinvariant.Theorem 11.15 (Analytic Dichotomy Lemma on Spanning). Let G be a connected,normed group (in the measure case a normed topological group). Suppose that an analyticset T ⊆ G spans a set of positive measure or a non-meagre set. Then T spansG.Proof. In the category case, the result follows from the Banach-Kuratowski Dichotomy,Th. 6.13 ([Ban-G, Satz 1], [Kur-1, Ch. VI. 13. XII], [Kel, Ch. 6 Prob. P p. 211]) by