Uniqueness of measure extensions in Banach spaces
Volume 175 / 2006
Abstract
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.
