In mathematics, the Pettis integral or Gelfand–Pettis integral, named after Israel M. Gelfand and Billy James Pettis, extends the definition of the Lebesgue integral to vector-valued functions on a measure space, by exploiting duality. The integral was introduced by Gelfand for the case when the measure space is an interval with Lebesgue measure. The integral is also called the weak integral in contrast to the Bochner integral, which is the strong integral.

Definition

Let f : X \to V where is a measure space and V is a topological vector space (TVS) with a continuous dual space V' that separates points (that is, if x \in V is nonzero then there is some l \in V' such that l(x) \neq 0), for example, V is a normed space or (more generally) is a Hausdorff locally convex TVS. Evaluation of a functional may be written as a duality pairing: The map f : X *to*** V is**** called**** **** if**** for all the scalar****-valued**** map **** is**** a measurable**** map.**** A weakly**** measurable**** map f : X *to*** V is**** said**** to**** be**** **** if**** there exists**** some**** e *in*** V such**** that**** for all the scalar****-valued**** map **** is**** Lebesgue**** integrable**** (that is****,**** ) and The map f : X *to*** V is**** said**** to**** be**** **** if**** for all **** and also**** for every **** there exists**** a vector**** eA *in*** V such**** that**** In**** this**** case****,**** eA is**** called**** the **** of**** f on**** A.**** Common**** notations for the Pettis**** integral**** e_A include To understand the motivation behind the definition of "weakly integrable", consider the special case where V is the underlying scalar field; that is, where V = \R or In this case, every linear functional \varphi on V is of the form for some scalar s \in V (that is, \varphi is just scalar multiplication by a constant), the condition simplifies to In particular, in this special case, f is weakly integrable on X if and only if f is Lebesgue integrable.

Relation to Dunford integral

The map f : X \to V is said to be **' if for all and also for every there exists a vector called the **' of f on A, such that where Identify every vector x \in V with the map scalar-valued functional on V' defined by This assignment induces a map called the canonical evaluation map and through it, V is identified as a vector subspace of the double dual V''. The space V is a semi-reflexive space if and only if this map is surjective. The f : X \to V is Pettis integrable if and only if d_A \in V for every

Properties

An immediate consequence of the definition is that Pettis integrals are compatible with continuous linear operators: If is linear and continuous and is Pettis integrable, then \Phi\circ f is Pettis integrable as well and The standard estimate for real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminorms and all Pettis integrable, holds. The right-hand side is the lower Lebesgue integral of a [0,\infty]-valued function, that is, Taking a lower Lebesgue integral is necessary because the integrand p\circ f may not be measurable. This follows from the Hahn-Banach theorem because for every vector v\in V there must be a continuous functional such that and for all w \in V,. Applying this to gives the result.

Mean value theorem

An important property is that the Pettis integral with respect to a finite measure is contained in the closure of the convex hull of the values scaled by the measure of the integration domain: This is a consequence of the Hahn-Banach theorem and generalizes the mean value theorem for integrals of real-valued functions: If V = \R, then closed convex sets are simply intervals and for, the following inequalities hold:

Existence

If V = \R^n is finite-dimensional then f is Pettis integrable if and only if each of f’s coordinates is Lebesgue integrable. If f is Pettis integrable and A\in\Sigma is a measurable subset of X, then by definition and are also Pettis integrable and If X is a topological space, its Borel-\sigma-algebra, \mu a Borel measure that assigns finite values to compact subsets, V is quasi-complete (that is, every bounded Cauchy net converges) and if f is continuous with compact support, then f is Pettis integrable. More generally: If f is weakly measurable and there exists a compact, convex and a null set such that, then f is Pettis-integrable.

Law of large numbers for Pettis-integrable random variables

Let be a probability space, and let V be a topological vector space with a dual space that separates points. Let be a sequence of Pettis-integrable random variables, and write for the Pettis integral of v_n (over X). Note that is a (non-random) vector in V, and is not a scalar value. Let denote the sample average. By linearity, \bar v_N is Pettis integrable, and Suppose that the partial sums converge absolutely in the topology of V, in the sense that all rearrangements of the sum converge to a single vector The weak law of large numbers implies that for every functional Consequently, in the weak topology on X. Without further assumptions, it is possible that does not converge to \lambda. To get strong convergence, more assumptions are necessary.