# Pettis integral

(Redirected from Gelfand–Pettis integral)

In mathematics, the Pettis integral or Gelfand–Pettis integral, named after I. M. Gelfand and B. J. 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

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

Choose any measurable set ${\displaystyle E\in \Sigma }$. We say that ${\displaystyle f}$ is Pettis integrable (over ${\displaystyle E}$) if there exists a vector ${\displaystyle e\in V}$ so that

${\displaystyle \langle \varphi ,e\rangle =\int _{E}\langle \varphi ,f(x)\rangle \,d\mu (x){\text{ for all functionals }}\varphi \in V^{*}.}$

In this case, we call ${\displaystyle e}$ the Pettis integral of ${\displaystyle f}$ (over ${\displaystyle E}$). Common notations for the Pettis integral ${\displaystyle e}$ include ${\displaystyle \int _{E}f\mu }$, ${\displaystyle \int _{E}f(t)\,d\mu (t)}$ and ${\displaystyle \mu [f1_{E}]}$.

A function is Pettis integrable (over ${\displaystyle X}$) if the scalar-valued function ${\displaystyle \varphi \circ f}$ is integrable for every functional ${\displaystyle \varphi \in X^{*}}$.

## Law of Large Numbers for Pettis integrable random variables

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

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

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

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