# Pettis integral

In mathematics, the Pettis integral or GelfandPettis 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 $(X,\Sigma ,\mu )$ is a measure space and $V$ is a topological vector space. Suppose that $V$ admits a dual space $V^{*}$ that separates points, e.g. $V$ is a Banach space or (more generally) is a locally-convex Hausdorff vector space. We write evaluation of a functional as duality pairing: $\langle \varphi ,x\rangle =\varphi [x]$ .

We say that $f$ is Pettis integrable if $\varphi \circ f\in L^{1}(X,\Sigma ,\mu )$ for all $\varphi \in V^{\ast }$ and there exists a vector $e\in V$ so that:

$\forall \varphi \in V^{*}:\qquad \langle \varphi ,e\rangle =\int _{X}\langle \varphi ,f(x)\rangle \,d\mu (x).$ In this case, we call $e$ the Pettis integral of $f$ . Common notations for the Pettis integral $e$ include

$\int _{X}fd\mu ,\qquad \int _{X}f(x)\,d\mu (x),\quad {\text{and}}\quad \mu [f].$ ## Properties

• An immediate consequence of the definition is that Pettis integrals are compatible with continuous, linear operators: If $\Phi :V_{1}\to V_{2}$ is and linear and continuous and $f:X\to V_{1}$ 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
$\left|\int _{X}f(x)\,d\mu (x)\right|\leqslant \int _{X}|f(x)|\,d\mu (x)$ for real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminorms $p:V\to \mathbb {R}$ and all Pettis integrable $f:X\to V,$ $p\left(\int _{X}f(x)\,d\mu (x)\right)\leqslant {\underline {\int _{X}}}p(f(x))\,d\mu (x)$ holds. The right hand side is the lower Lebesgue integral of a $[0,\infty ]$ -valued function, i.e.
${\underline {\int _{X}}}g\,d\mu :=\sup \left\{\left.\int _{X}h\,d\mu \right|h:X\to [0,\infty ]{\text{ is measurable and }}0\leqslant h\leqslant g\right\}.$ 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 $\varphi \in V^{\ast }$ such that $\varphi (v)=p(v)$ and $\forall w\in V:|\varphi (w)|\leq p(w)$ . Applying this to $v:=\int _{X}f\,d\mu$ 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:

$\mu (A)<\infty \implies \int _{A}f\,d\mu \in \mu (A)\cdot {\overline {co(f(A))}}$ This is a consequence of the Hahn-Banach theorem and generalises the mean value theorem for integrals of real-valued functions: If $V=\mathbb {R} ,$ then closed convex sets are simply intervals and for $f:X\to [a,b],$ the inequalities

$\mu (A)a\leqslant \int _{A}f\,d\mu \leqslant \mu (A)b$ hold.

### 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 $f_{|A}:A\to V$ and $f\cdot 1_{A}:X\to V$ 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, $\Sigma ={\mathfrak {B}}_{X}$ its Borel-$\sigma$ -algebra, $\mu$ a Borel measure that assigns finite values to compact subsets, $V$ is quasi-complete (i.e. if 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 $C\subseteq V$ 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 $v_{n}\colon \Omega \to V$ 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 the weak topology on $X$ .

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.