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.
Let where is a measure space and is a topological vector space (TVS) with a continuous dual space that separates points (i.e. if is nonzero then there is some such that ), e.g. 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 is called weakly measurable if for all , the scalar-valued map is a measurable map.
A weakly measurable map is said to be weakly integrable on if there exists some such that for all , the scalar-valued map is Lebesgue integrable (that is, ) and
The map is said to be Pettis integrable if for all and also for every there exists a vector such that for all
In this case, we call the Pettis integral of on Common notations for the Pettis integral include
To understand the motivation behind the definition of "weakly integrable", consider the special case where is the underlying scalar field; that is, where or In this case, every linear functional on is of the form for some scalar (that is, is just scalar multiplication by a constant), the condition
In particular, in this special case, is weakly integrable on if and only if is Lebesgue integrable.
- An immediate consequence of the definition is that Pettis integrals are compatible with continuous, linear operators: If is and linear and continuous and is Pettis integrable, then 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 -valued function, i.e. Taking a lower Lebesgue integral is necessary because the integrand may not be measurable. This follows from the Hahn-Banach theorem because for every vector there must be a continuous functional such that and for all , . Applying this to it 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 generalises the mean value theorem for integrals of real-valued functions: If , then closed convex sets are simply intervals and for , the inequalities
- If is finite-dimensional then is Pettis integrable if and only if each of 's coordinates is Lebesgue integrable.
- If is Pettis integrable and is a measurable subset of , then by definition and are also Pettis integrable and
- If is a topological space, its Borel--algebra, a Borel measure that assigns finite values to compact subsets, is quasi-complete (i.e. every bounded Cauchy net converges) and if is continuous with compact support, then is Pettis integrable.
- More generally: If is weakly measurable and there exists a compact, convex and a null set such that , then is Pettis-integrable.
Law of large numbers for Pettis-integrable random variables
Let be a probability space, and let 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 (over ). Note that is a (non-random) vector in , and is not a scalar value.
denote the sample average. By linearity, is Pettis integrable, and
Suppose that the partial sums
converge absolutely in the topology of , 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 .
Without further assumptions, it is possible that does not converge to . To get strong convergence, more assumptions are necessary.