※ 引述《Rhymer (Design)》之铭言： : Let (Ω,Σ) be a measurable space and X a Banach space. : f:Ω->X is called weakly measurable if for each linear functional : x' in X'(the norm dual of X), : the scalar function < f, x' > is measurable. : 类比於 storngly measurable 的定义, 我猜想 : f is weakly measurable iff there exists a sequence of step functions : φ_n:Ω->X such that < φ_n, x' > converges to < f, x' > for each x' in X'. : 不知道这个猜测是否正确? : <= 方向的证明很简单, 但是 => 却没有头绪... : 恳请板上神人指点迷津! : 感谢!! 结果是否定的. 如果 φ_n -> f weakly a.e. 则 f 为 storngly measurable. 推论其实很简单, 关键是 "Pettis' theorem" 以及 "subspace 的 weak closure 与 norm closure 是相同的" 这两件事. 可惜我想不到... 以下是来自 Mathoverflow 的解答: http://tinyurl.com/3jy9rzh (Answered by Michael Renardy. Rephrased.) If φ_n converges to f weakly a.e., then the range of f is contained in a separable subspace of X (注). However, if f is weakly measurable, then by Pettis' theorem, which states that f is strongly measurable iff it is weakly measurable and its range is almost separable, f is strongly measurable. (注) By Mazur's theorem, the closure and weak closure of a subspace in a normed space is the same. See An introduction to Banach space theory corollary 2.5.17 by Robert E. Megginson. --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 220.133.4.190