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
where
is a measure space and
is a topological vector space (TVS) with a continuous dual space
that separates points (that is, if
is nonzero then there is some
such that
), for example,
is a normed space or (more generally) is a Hausdorff locally convex TVS. Evaluation of a functional may be written as a duality pairing:
![{\displaystyle \langle \varphi ,x\rangle =\varphi [x].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92584443166736d482e32139463a5eddea49c1d2)
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
![{\displaystyle \varphi (e)=\int _{X}\varphi (f(x))\,\mathrm {d} \mu (x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34a54bb38706e0baffcc98767a03121f8e8f752c)
The map
is said to be Pettis integrable if
for all
and also for every
there exists a vector
such that
![{\displaystyle \langle \varphi ,e_{A}\rangle =\int _{A}\langle \varphi ,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/623063d5f1d8d740d6ad404f004d13d58251a90b)
In this case,
is called the Pettis integral of
on
Common notations for the Pettis integral
include
![{\displaystyle \int _{A}f\,\mathrm {d} \mu ,\qquad \int _{A}f(x)\,\mathrm {d} \mu (x),\quad {\text{and, in case that}}~A=X~{\text{is understood,}}\quad \mu [f].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df9eec809a28e7d6c6e2e4a307ee58fbbbb1fbdc)
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
![{\displaystyle \varphi (e)=\int _{A}\varphi (f(x))\,\mathrm {d} \mu (x)\quad {\text{for all}}~\varphi \in V',}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b970bac259866e5e5a213df2c2797e4c226f0e9)
simplifies to
![{\displaystyle se=\int _{A}sf(x)\,\mathrm {d} \mu (x)\quad {\text{for all scalars}}~s.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a11c985ae5613867844729cc88999057edc05220)
In particular, in this special case,
![{\displaystyle f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61)
is weakly integrable on
![{\displaystyle X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
if and only if
![{\displaystyle f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61)
is Lebesgue integrable.
Relation to Dunford integral
The map
is said to be Dunford integrable if
for all
and also for every
there exists a vector
called the Dunford integral of
on
such that
![{\displaystyle \langle d_{A},\varphi \rangle =\int _{A}\langle \varphi ,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e56193647af6781df31a7acdb0229cd5c39934d)
where
Identify every vector
with the map scalar-valued functional on
defined by
This assignment induces a map called the canonical evaluation map and through it,
is identified as a vector subspace of the double dual
The space
is a semi-reflexive space if and only if this map is surjective. The
is Pettis integrable if and only if
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
is Pettis integrable as well and
![{\displaystyle \int _{X}\Phi (f(x))\,d\mu (x)=\Phi \left(\int _{X}f(x)\,d\mu (x)\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bb56048b98ce2bcea9c2b4aafb1689b0a86baf9)
The standard estimate
![{\displaystyle \left|\int _{X}f(x)\,d\mu (x)\right|\leq \int _{X}|f(x)|\,d\mu (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/201ba5eecee458b7c1f6e7f13c62acf923bee863)
for real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminorms
![{\displaystyle p\colon V\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b8a7db7fed03868169ffc711bd742c98afa5976)
and all Pettis integrable
![{\displaystyle f\colon X\to V}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51bce2b9cd4724fae16883cb99afc6aa9d84e8ef)
,
![{\displaystyle p\left(\int _{X}f(x)\,d\mu (x)\right)\leq {\underline {\int _{X}}}p(f(x))\,d\mu (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbeeca2632ae46b902df971b66d7046fdb9e8f8e)
holds. The right-hand side is the lower Lebesgue integral of a
![{\displaystyle [0,\infty ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52088d5605716e18068a460dec118214954a68e9)
-valued function, that is,
![{\displaystyle {\underline {\int _{X}}}g\,d\mu :=\sup \left\{\left.\int _{X}h\,d\mu \;\right|\;h\colon X\to [0,\infty ]{\text{ is measurable and }}0\leq h\leq g\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7131bfd237f02e8e06f56c05225a15da5c3bafa)
Taking a lower Lebesgue integral is necessary because the integrand
![{\displaystyle p\circ f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d28e34f4c6ecfb577a55bce6e3cee686945bb2f4)
may not be measurable. This follows from the Hahn-Banach theorem because for every vector
![{\displaystyle v\in V}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99886ebbde63daa0224fb9bf56fa11b3c8a6f4fb)
there must be a continuous functional
![{\displaystyle \varphi \in V^{*}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cb13328d30ab64d68a901fea1e99b0d789f983d)
such that
![{\displaystyle \varphi (v)=p(v)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5652e17d5dd4b7cba2ac5df5dc9bae9131a1324e)
and for all
![{\displaystyle w\in V}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb3ba2e494febb4b85886a94ea45400bbfa30176)
,
![{\displaystyle |\varphi (w)|\leq p(w)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08a7cd2afc3aad18ae645060222904e3d21a4766)
. Applying this to
![{\displaystyle v:=\int _{X}f\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1aa888efa5471aa673d6a47540932b9df9cb05f0)
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:
![{\displaystyle \mu (A)<\infty {\text{ implies }}\int _{A}f\,d\mu \in \mu (A)\cdot {\overline {\operatorname {co} (f(A))}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/233429f7868a258f96ca26fd5d0ae5a91906e0aa)
This is a consequence of the Hahn-Banach theorem and generalizes the mean value theorem for integrals of real-valued functions: If
, then closed convex sets are simply intervals and for
, the following inequalities hold:
![{\displaystyle \mu (A)a~\leq ~\int _{A}f\,d\mu ~\leq ~\mu (A)b.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f7d3af78b396aef7769fa93bbee8b012e038496)
Existence
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
![{\displaystyle \int _{A}f_{|A}\,d\mu =\int _{X}f\cdot 1_{A}\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3655743f719e4b417b52085cd18aaa5111976f0)
If
is a topological space,
its Borel-
-algebra,
a Borel measure that assigns finite values to compact subsets,
is quasi-complete (that is, 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.
Let
![{\displaystyle {\bar {v}}_{N}:={\frac {1}{N}}\sum _{n=1}^{N}v_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffc67978d1344019e7fe4ac21351494d462dd2e3)
denote the sample average. By linearity,
![{\displaystyle {\bar {v}}_{N}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6155309de37d39ef097a9e8e45f2bf509d55789)
is Pettis integrable, and
![{\displaystyle \operatorname {E} [{\bar {v}}_{N}]={\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [v_{n}]\in V.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecf9cb2d6a7643bc95c6999887f0f5816c9737db)
Suppose that the partial sums
![{\displaystyle {\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [{\bar {v}}_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48a2ef26625246c8cc99949aa469f9fdee2e085c)
converge absolutely in the topology of
![{\displaystyle V,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ace9595e3ce66fdec7e9d30202626accd676b11e)
in the sense that all rearrangements of the sum converge to a single vector
![{\displaystyle \lambda \in V.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/642724eb37c41bc63b501532f37fa3fb93160090)
The weak law of large numbers implies that
![{\displaystyle \langle \varphi ,\operatorname {E} [{\bar {v}}_{N}]-\lambda \rangle \to 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56671ce061d998074a6fca038c2c78fce886b854)
for every functional
![{\displaystyle \varphi \in V^{*}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01f664dcce2521e039d19b835d5c037596ec1d88)
Consequently,
![{\displaystyle \operatorname {E} [{\bar {v}}_{N}]\to \lambda }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3ed2a98fe8adf477397c8140daf38006bd7c336)
in the
weak topology on
Without further assumptions, it is possible that
does not converge to
[citation needed] To get strong convergence, more assumptions are necessary.[citation needed]
See also
References
- James K. Brooks, Representations of weak and strong integrals in Banach spaces, Proceedings of the National Academy of Sciences of the United States of America 63, 1969, 266–270. Fulltext MR0274697
- 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, 35–40 Zbl 0014.16202
- Michel Talagrand, Pettis Integral and Measure Theory, Memoirs of the AMS no. 307 (1984) MR0756174
- Sobolev, V. I. (2001) [1994], "Pettis integral", Encyclopedia of Mathematics, EMS Press
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