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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Lebesgue's density theorem ： ウィキペディア英語版 Lebesgue's density theorem In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set $A\backslash subset\; \backslash R^n$, the "density" of ''A'' is 0 or 1 at almost every point in $\backslash R^n$. Additionally, the "density" of ''A'' is 1 at almost every point in ''A''. Intuitively, this means that the "edge" of ''A'', the set of points in ''A'' whose "neighborhood" is partially in ''A'' and partially outside of ''A'', is negligible. Let μ be the Lebesgue measure on the Euclidean space R''n'' and ''A'' be a Lebesgue measurable subset of R''n''. Define the approximate density of ''A'' in a ε-neighborhood of a point ''x'' in R''n'' as :$d\_\backslash varepsilon(x)=\backslash frac$ where ''B''ε denotes the closed ball of radius ε centered at ''x''. Lebesgue's density theorem asserts that for almost every point ''x'' of ''A'' the density :$d(x)=\backslash lim\_\; d\_(x)$ exists and is equal to 1. In other words, for every measurable set ''A'', the density of ''A'' is 0 or 1 almost everywhere in R''n''. However, it is a curious fact that if μ(''A'') > 0 and , then there are always points of R''n'' where the density is neither 0 nor 1. For example, given a square in the plane, the density at every point inside the square is 1, on the edges is 1/2, and at the corners is 1/4. The set of points in the plane at which the density is neither 0 nor 1 is non-empty (the square boundary), but it is negligible. The Lebesgue density theorem is a particular case of the Lebesgue differentiation theorem. Thus, this theorem is also true for every finite Borel measure on R''n'' instead of Lebesgue measure, see Discussion. == See also == * Lebesgue differentiation theorem. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Lebesgue's density theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.