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On the convergence of Σ∞n = 1f(nx) for measurable functions
Published online by Cambridge University Press: 26 February 2010
Abstract
Questions of Haight and of Weizsäcker are answered in the following result. There exists a measurable function f: (0, + ∞) → {0,1} and two non-empty intervals IFI∞⊂[½,1) such that Σ∞n = 1f(nx) = +∞ for everyx εI∞, and Σ∞n = 1f(nx) >+∞ for almost every xεIf. The function f may be taken to be the characteristic function of an open set E.
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- Copyright © University College London 1999
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