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On the integration of vector-valued functions. (English) Zbl 0790.28004
We investigate methods of integrating functions from \([0,1]\) to a Banach space, in particular the McShane integral as defined by R. A. Gordon [Ill. J. Math. 34, No. 3, 557-567 (1990; Zbl 0714.28008)], and what we call the ‘Talagrand integral’, based on ideas from M. Talagrand [Ann. Probab. 15, 837-870 (1987; Zbl 0632.60024)]. Briefly: if \(X\) is a Banach space, a function \(\phi: [0,1]\to X\) is McShane integrable, with integral \(w\in X\), if for every \(\varepsilon> 0\) there is a function \(\delta: [0,1]\to ]0,\infty[\) such that \(\| w- \sum^ n_{i=1} (a_ i- a_{i-1})\phi(t_ i)\|\leq\varepsilon\) whenever \(0= a_ 0\leq a_ 1\leq\cdots\leq a_ n= 1\) and \(t_ 1,\dots,t_ n\in [0,1]\) are such that \(t_ i- \delta(t_ i)\leq a_{i-1}\leq a_ i\leq t_ i+ \delta(t_ i)\) for every \(i\leq n\); and \(\phi\) is Talagrand integrable, with integral \(w\), if \(\lim_{n\to\infty}{1\over n}\sum^ n_{i=1}\phi(t_ i)= w\) for almost all sequences \(t_ 1,t_ 2,\dots\) in \([0,1]\).
It is known that a Bochner integrable function is Talagrand integrable and McShane integrable, and that a Talagrand integrable function is Pettis integrable; we show that a McShane integrable function is Pettis integrable. We show there is a Talagrand integrable function which is not McShane integrable and that there is a McShane integrable function which is not Talagrand integrable. If the unit ball of the dual \(X^*\) of \(X\) is separable, however, then every McShane integrable function from \([0,1]\) to \(X\) must be Talagrand integrable, but need not be Bochner integrable. If \(X\) itself is separable, then every Talagrand integrable function is Bochner integrable, and every Pettis integrable function is McShane integrable.
We give a convergence theorem for the McShane integral, as follows: if \(\langle\phi\rangle_{n\in\mathbb{N}}\) is a sequence of McShane integrable functions from \([0,1]\) to \(X\), and if (i) \(\phi(t)=\lim_{n\to\infty}\phi_ n(t)\) exists in \(X\) (for the norm topology) for every \(t\in [0,1]\) and (ii) \(\nu E=\lim_{n\to\infty}\int_ E\phi_ n\) exists in \(X\) (for the weak topology) for every measurable set \(E\subseteq [0,1]\), then \(\phi\) is McShane integrable and \(\nu\) is the indefinite Pettis integral of \(\phi\).

MSC:
28B05 Vector-valued set functions, measures and integrals
46G10 Vector-valued measures and integration
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