Letwhereis a measure space andis atopological vector space(TVS) with a continuous dual spacethat separates points (that is, ifis nonzero then there is somesuch that), for example,is anormed spaceor (more generally) is a Hausdorfflocally convexTVS.
Evaluation of a functional may be written as aduality pairing:
The mapis calledweakly measurableif for allthe scalar-valued mapis ameasurable map.
A weakly measurable mapis said to beweakly integrableonif there exists somesuch that for allthe scalar-valued mapisLebesgue integrable(that is,) and
The mapis said to bePettis integrableiffor alland also for everythere exists a vectorsuch that
In this case,is called thePettis integralofonCommon notations for the Pettis integralinclude
To understand the motivation behind the definition of "weakly integrable", consider the special case whereis the underlying scalar field; that is, whereorIn this case, every linear functionalonis of the formfor some scalar(that is,is just scalar multiplication by a constant), the condition
simplifies to
In particular, in this special case,is weakly integrable onif and only ifis Lebesgue integrable.
The mapis said to beDunford integrableiffor alland also for everythere exists a vectorcalled theDunford integralofonsuch that
where
Identify every vectorwith the map scalar-valued functional ondefined byThis assignment induces a map called the canonical evaluation map and through it,is identified as a vector subspace of the double dual
The spaceis asemi-reflexive spaceif and only if this map issurjective.
Theis Pettis integrable if and only iffor every
An immediate consequence of the definition is that Pettis integrals are compatible with continuous linear operators: Ifis linear and continuous andis Pettis integrable, thenis Pettis integrable as well and
The standard estimatefor real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminormsand all Pettis integrable,holds. The right-hand side is the lower Lebesgue integral of a-valued function, that is,Taking a lower Lebesgue integral is necessary because the integrandmay not be measurable. This follows from theHahn-Banach theorembecause for every vectorthere must be a continuous functionalsuch thatand for all,.Applying this togives the result.
An important property is that the Pettis integral with respect to a finite measure is contained in the closure of theconvex hullof the values scaled by the measure of the integration domain:
Ifis finite-dimensional thenis Pettis integrable if and only if each of’s coordinates is Lebesgue integrable.
Ifis Pettis integrable andis a measurable subset of,then by definitionandare also Pettis integrable and
Ifis a topological space,itsBorel--algebra,aBorel measurethat assigns finite values to compact subsets,isquasi-complete(that is, everyboundedCauchy netconverges) and ifis continuous with compact support, thenis Pettis integrable.
More generally: Ifis weakly measurable and there exists a compact, convexand a null setsuch that,thenis Pettis-integrable.
Law of large numbers for Pettis-integrable random variables
Letbe a probability space, and letbe a topological vector space with a dual space that separates points. Letbe a sequence of Pettis-integrable random variables, and writefor the Pettis integral of(over). Note thatis a (non-random) vector inand is not a scalar value.
Let
denote the sample average. By linearity,is Pettis integrable, and
Suppose that the partial sums
converge absolutely in the topology ofin the sense that all rearrangements of the sum converge to a single vectorThe weak law of large numbers implies thatfor every functionalConsequently,in theweak topologyon
Without further assumptions, it is possible thatdoes not converge to[citation needed]To get strong convergence, more assumptions are necessary.[citation needed]
Israel M. Gel'fand,Sur un lemme de la théorie des espaces linéaires,Commun. Inst. Sci. Math. et Mecan., Univ. Kharkoff et Soc. Math. Kharkoff, IV. Ser. 13, 1936, 35–40Zbl0014.16202