Bourbaki

习题 13, § 2

标签 09BL,永久不变。引用为 [Bourbaki, Tag 09BL]。

积分, VI, p. 65

Bourbaki marks this one as harder than the rest.

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Let GG be a Hausdorff barreled space, F=G′F = G' its dual; if, when FF is equipped with a topology compatible with the duality between FF and GG, m\mathbf{m} is a vectorial measure on TT with values in FF, then m\mathbf{m} is again a vectorial measure when FF is equipped with the strong topology of the dual of GG.

假设 m\mathbf{m} 是基 μ\mu 的标量;那么当FF配备拓扑τ(F,G)\tau(F, G)时,m\mathbf{m}相对于μ\mu绝对连续(Exer. 12 c))。 In order that m\mathbf{m} be absolutely continuous with respect to μ\mu when FF is equipped with the strong topology, it is necessary and sufficient that, for every compact subset KK of TT and every sequence (An)(A_n) of μ\mu-measurable subsets of KK, the image BB under m\mathbf{m} of the closure in L1(μ)\mathscr{L}^1(\mu) of the set of step functions of absolute value ≤1\leq 1, over the clan generated by the sequence of the AnA_n (Ch. IV, §4, No. 9),包含是G′G'的强拓扑的稠密可数子集。 (要看到条件就足够了,约化对于T=KT = K是石头空间的情况(第1章) IV, §4,埃克瑟。 10)。 然后通过反证法论证:如果(An)(A_n)是KK的子集的序列,并且既开又闭,我们可以假设,在经过上到KK的可度量商空间K1K_1(Exer.10 a)),连续K1K_1相应的函数到φAn\varphi_{A_n}形式总共集合C(K1)\mathscr{C}(K_1)。 另一方面,考虑到GG除以与BB正交的子空间的商,可以假设对于拓扑σ(G′,G)\sigma(G', G),BB是F=G′F = G'的总和;如果HH是由BB生成的F=G′F = G'的子空间,假设则意味着GG的每个有界子集CC对于σ(G,H)\sigma(G, H)都是预紧且可度量的。 因此,GG的点的每个有界序列(zn)(z_n)都有一个σ(G,H)\sigma(G, H)的子序列柯西;证明如果(zn)(z_n)是σ(G,H)\sigma(G, H)的柯西序列,并且如果写成zn∘m=gzn⋅μz_n \circ \mathbf{m} = g_{z_n} \cdot \mu,则L1(μ)L^1(\mu)中的gzng_{z_n}的类的序列是拓扑的柯西序列 σ(L1,L∞)\sigma(L^1, L^\infty);最后,结论通过使用Exers与此相矛盾。第 15 章和第 16 章 V, §5。)

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Google 翻译生成的粗略草稿,质量较低。术语依照术语表,但句子可能有误;之后会由仔细的译文取代。