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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