Uniqueness of measure extensions in Banach spaces

J. Rodríguez¹ ; G. Vera¹

¹Departamento de Matemáticas Universidad de Murcia 30100 Espinardo (Murcia), Spain

Studia Mathematica, Tome 175 (2006) no. 2, pp. 139-155

Résumé

Let $X$ be a Banach space, $B\subset B_{X^{*}}$ a norming set and $\mathfrak{T}(X,B)$ the topology on $X$ of pointwise convergence on $B$. We study the following question: given two (non-negative, countably additive and finite) measures $\mu_{1}$ and $\mu_{2}$ on ${\rm Baire}(X,w)$ which coincide on ${\rm Baire}(X,\mathfrak{T}(X,B))$, does it follow that $\mu_{1}=\mu_{2}$? It turns out that this is not true in general, although the answer is affirmative provided that both $\mu_{1}$ and $\mu_{2}$ are convexly $\tau$-additive (e.g. when $X$ has the Pettis Integral Property). For a Banach space $Y$ not containing isomorphic copies of $\ell^{1}$, we show that $Y^{*}$ has the Pettis Integral Property if and only if every measure on ${\rm Baire}(Y^{*},w^{*})$ admits a unique extension to ${\rm Baire}(Y^{*},w)$. We also discuss the coincidence of the two $\sigma$-algebras involved in such results. Some other applications are given.

Détail

DOI : 10.4064/sm175-2-3

Keywords: banach space subset * norming set mathfrak topology nbsp pointwise convergence nbsp study following question given non negative countably additive finite measures nbsp nbsp baire which coincide nbsp baire mathfrak does follow turns out general although answer affirmative provided nbsp convexly tau additive has pettis integral property banach space nbsp containing isomorphic copies nbsp ell * has pettis integral property only every measure baire * * admits unique extension nbsp baire * discuss coincidence sigma algebras involved results other applications given