Pettis integral

Inmathematics,thePettis integralorGelfand–Pettis integral,named afterIsrael M. GelfandandBilly James Pettis,extends the definition of theLebesgue integralto vector-valued functions on ameasure space,by exploitingduality.The integral was introduced by Gelfand for the case when the measure space is an interval withLebesgue measure.The integral is also called theweak integralin contrast to theBochner integral,which is the strong integral.

Definition

Let$f:X\to V$where$(X,\Sigma,\mu )$is a measure space and$V$is atopological 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 anormed spaceor (more generally) is a Hausdorfflocally convexTVS. Evaluation of a functional may be written as aduality pairing: $$\langle \varphi,x\rangle =\varphi [x].$$

The map$f:X\to V$is calledweakly measurableif for all$\varphi \in V',$the scalar-valued map$\varphi \circ f$is ameasurable map. A weakly measurable map$f:X\to V$is said to beweakly integrableon$X$if there exists some$e\in V$such that for all$\varphi \in V',$the scalar-valued map$\varphi \circ f$isLebesgue integrable(that is,$\varphi \circ f\in L^{1}\left(X,\Sigma,\mu \right)$) and $$\varphi (e)=\int _{X}\varphi (f(x))\,\mathrm {d} \mu (x).$$

The map$f:X\to V$is said to bePettis integrableif$\varphi \circ f\in L^{1}\left(X,\Sigma,\mu \right)$for all$\varphi \in V^{\prime }$and also for every$A\in \Sigma$there exists a vector$e_{A}\in V$such that $$\langle \varphi,e_{A}\rangle =\int _{A}\langle \varphi,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'.$$

In this case,$e_{A}$is called thePettis integralof$f$on$A.$Common notations for the Pettis integral$e_{A}$include $$\int _{A}f\,\mathrm {d} \mu,\qquad \int _{A}f(x)\,\mathrm {d} \mu (x),\quad {\text{and, in case that}}~A=X~{\text{is understood,}}\quad \mu [f].$$

To understand the motivation behind the definition of "weakly integrable", consider the special case where$V$is the underlying scalar field; that is, where$V=\mathbb {R}$or$V=\mathbb {C}.$In this case, every linear functional$\varphi$on$V$is of the form$\varphi (y)=sy$for some scalar$s\in V$(that is,$\varphi$is just scalar multiplication by a constant), the condition $$\varphi (e)=\int _{A}\varphi (f(x))\,\mathrm {d} \mu (x)\quad {\text{for all}}~\varphi \in V',$$ simplifies to $$se=\int _{A}sf(x)\,\mathrm {d} \mu (x)\quad {\text{for all scalars}}~s.$$ 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 beDunford integrableif$\varphi \circ f\in L^{1}\left(X,\Sigma,\mu \right)$for all$\varphi \in V^{\prime }$and also for every$A\in \Sigma$there exists a vector$d_{A}\in V'',$called theDunford integralof$f$on$A,$such that $$\langle d_{A},\varphi \rangle =\int _{A}\langle \varphi,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'$$ where$\langle d_{A},\varphi \rangle =d_{A}(\varphi ).$

Identify every vector$x\in V$with the map scalar-valued functional on$V'$defined by$\varphi \in V'\mapsto \varphi (x).$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 asemi-reflexive spaceif and only if this map issurjective. The$f:X\to V$is Pettis integrable if and only if$d_{A}\in V$for every$A\in \Sigma.$

Properties

An immediate consequence of the definition is that Pettis integrals are compatible with continuous linear operators: If$\Phi \colon V_{1}\to V_{2}$is linear and continuous and$f\colon X\to V_{1}$is Pettis integrable, then$\Phi \circ f$is Pettis integrable as well and$$\int _{X}\Phi (f(x))\,d\mu (x)=\Phi \left(\int _{X}f(x)\,d\mu (x)\right).$$

The standard estimate$$\left|\int _{X}f(x)\,d\mu (x)\right|\leq \int _{X}|f(x)|\,d\mu (x)$$for real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminorms$p\colon V\to \mathbb {R}$and all Pettis integrable$f\colon X\to V$,$$p\left(\int _{X}f(x)\,d\mu (x)\right)\leq {\underline {\int _{X}}}p(f(x))\,d\mu (x)$$holds. The right-hand side is the lower Lebesgue integral of a$[0,\infty ]$-valued function, that is,$${\underline {\int _{X}}}g\,d\mu:=\sup \left\{\left.\int _{X}h\,d\mu \;\right|\;h\colon X\to [0,\infty ]{\text{ is measurable and }}0\leq h\leq g\right\}.$$Taking a lower Lebesgue integral is necessary because the integrand$p\circ f$may not be measurable. This follows from theHahn-Banach theorembecause for every vector$v\in V$there must be a continuous functional$\varphi \in V^{*}$such that$\varphi (v)=p(v)$and for all$w\in V$,$|\varphi (w)|\leq p(w)$.Applying this to$v:=\int _{X}f\,d\mu$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 theconvex hullof the values scaled by the measure of the integration domain: $$\mu (A)<\infty {\text{ implies }}\int _{A}f\,d\mu \in \mu (A)\cdot {\overline {\operatorname {co} (f(A))}}$$

This is a consequence of theHahn-Banach theoremand generalizes themean value theorem for integrals of real-valued functions:If$V=\mathbb {R}$,then closed convex sets are simply intervals and for$f\colon X\to [a,b]$,the following inequalities hold: $$\mu (A)a~\leq ~\int _{A}f\,d\mu ~\leq ~\mu (A)b.$$

Existence

If$V=\mathbb {R} ^{n}$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$f_{|A}\colon A\to V$and$f\cdot 1_{A}\colon X\to V$are also Pettis integrable and$$\int _{A}f_{|A}\,d\mu =\int _{X}f\cdot 1_{A}\,d\mu.$$

If$X$is a topological space,$\Sigma ={\mathfrak {B}}_{X}$itsBorel-$\sigma$-algebra,$\mu$aBorel measurethat assigns finite values to compact subsets,$V$isquasi-complete(that is, everyboundedCauchy netconverges) 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$C\subseteq V$and a null set$N\subseteq X$such that$f(X\setminus N)\subseteq C$,then$f$is Pettis-integrable.

Law of large numbers for Pettis-integrable random variables

Let$(\Omega,{\mathcal {F}},\operatorname {P} )$be a probability space, and let$V$be a topological vector space with a dual space that separates points. Let$v_{n}:\Omega \to V$be a sequence of Pettis-integrable random variables, and write$\operatorname {E} [v_{n}]$for the Pettis integral of$v_{n}$(over$X$). Note that$\operatorname {E} [v_{n}]$is a (non-random) vector in$V,$and is not a scalar value.

Let $${\bar {v}}_{N}:={\frac {1}{N}}\sum _{n=1}^{N}v_{n}$$ denote the sample average. By linearity,${\bar {v}}_{N}$is Pettis integrable, and $$\operatorname {E} [{\bar {v}}_{N}]={\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [v_{n}]\in V.$$

Suppose that the partial sums $${\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [{\bar {v}}_{n}]$$ converge absolutely in the topology of$V,$in the sense that all rearrangements of the sum converge to a single vector$\lambda \in V.$The weak law of large numbers implies that$\langle \varphi,\operatorname {E} [{\bar {v}}_{N}]-\lambda \rangle \to 0$for every functional$\varphi \in V^{*}.$Consequently,$\operatorname {E} [{\bar {v}}_{N}]\to \lambda$in theweak topologyon$X.$

Without further assumptions, it is possible that$\operatorname {E} [{\bar {v}}_{N}]$does not converge to$\lambda.$To get strong convergence, more assumptions are necessary.

See also

• Bochner measurable function
• Bochner integral– generalization of the Lebesgue integral to Banach-space valued functionsPages displaying wikidata descriptions as a fallback
• Bochner space– Type of topological space
• Vector measure
• Weakly measurable function

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 America63, 1969, 266–270.FulltextMR0274697
• 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
• 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