Pettis integral

Let ${\displaystyle f:X\to V}$ where ${\displaystyle (X,\Sigma ,\mu )}$ is a measure space and ${\displaystyle V}$ is a topological vector space (TVS) with a continuous dual space ${\displaystyle V'}$ that separates points (that is, if ${\displaystyle x\in V}$ is nonzero then there is some ${\displaystyle l\in V'}$ such that ${\displaystyle l(x)\neq 0}$), for example, ${\displaystyle V}$ is a normed space or (more generally) is a Hausdorff locally convex TVS. Evaluation of a functional may be written as a duality pairing:

$${\displaystyle \langle \varphi ,x\rangle =\varphi [x].}$$

The map ${\displaystyle f:X\to V}$ is called weakly measurable if for all ${\displaystyle \varphi \in V',}$ the scalar-valued map ${\displaystyle \varphi \circ f}$ is a measurable map. A weakly measurable map ${\displaystyle f:X\to V}$ is said to be weakly integrable on ${\displaystyle X}$ if there exists some ${\displaystyle e\in V}$ such that for all ${\displaystyle \varphi \in V',}$ the scalar-valued map ${\displaystyle \varphi \circ f}$ is Lebesgue integrable (that is, ${\displaystyle \varphi \circ f\in L^{1}\left(X,\Sigma ,\mu \right)}$) and

$${\displaystyle \varphi (e)=\int _{X}\varphi (f(x))\,\mathrm {d} \mu (x).}$$

The map ${\displaystyle f:X\to V}$ is said to be Pettis integrable if ${\displaystyle \varphi \circ f\in L^{1}\left(X,\Sigma ,\mu \right)}$ for all ${\displaystyle \varphi \in V^{\prime }}$ and also for every ${\displaystyle A\in \Sigma }$ there exists a vector ${\displaystyle e_{A}\in V}$ such that

$${\displaystyle \langle \varphi ,e_{A}\rangle =\int _{A}\langle \varphi ,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'.}$$

In this case, ${\displaystyle e_{A}}$ is called the Pettis integral of ${\displaystyle f}$ on ${\displaystyle A.}$ Common notations for the Pettis integral ${\displaystyle e_{A}}$ include

$${\displaystyle \int _{A}f\,\mathrm {d} \mu ,\qquad \int _{A}f(x)\,\mathrm {d} \mu (x),\quad {\text{and, in case that}}~A=X~{\text{is understood,}}\quad \mu [f].}$$

To understand the motivation behind the definition of "weakly integrable", consider the special case where ${\displaystyle V}$ is the underlying scalar field; that is, where ${\displaystyle V=\mathbb {R} }$ or ${\displaystyle V=\mathbb {C} .}$ In this case, every linear functional ${\displaystyle \varphi }$ on ${\displaystyle V}$ is of the form ${\displaystyle \varphi (y)=sy}$ for some scalar ${\displaystyle s\in V}$ (that is, ${\displaystyle \varphi }$ is just scalar multiplication by a constant), the condition

$${\displaystyle \varphi (e)=\int _{A}\varphi (f(x))\,\mathrm {d} \mu (x)\quad {\text{for all}}~\varphi \in V',}$$

simplifies to

$${\displaystyle se=\int _{A}sf(x)\,\mathrm {d} \mu (x)\quad {\text{for all scalars}}~s.}$$

In particular, in this special case, ${\displaystyle f}$ is weakly integrable on ${\displaystyle X}$ if and only if ${\displaystyle f}$ is Lebesgue integrable.

The map ${\displaystyle f:X\to V}$ is said to be Dunford integrable if ${\displaystyle \varphi \circ f\in L^{1}\left(X,\Sigma ,\mu \right)}$ for all ${\displaystyle \varphi \in V^{\prime }}$ and also for every ${\displaystyle A\in \Sigma }$ there exists a vector ${\displaystyle d_{A}\in V'',}$ called the Dunford integral of ${\displaystyle f}$ on ${\displaystyle A,}$ such that

$${\displaystyle \langle d_{A},\varphi \rangle =\int _{A}\langle \varphi ,f(x)\rangle \,\mathrm {d} \mu (x)\quad {\text{ for all }}\varphi \in V'}$$

where ${\displaystyle \langle d_{A},\varphi \rangle =d_{A}(\varphi ).}$

Identify every vector ${\displaystyle x\in V}$ with the map scalar-valued functional on ${\displaystyle V'}$ defined by ${\displaystyle \varphi \in V'\mapsto \varphi (x).}$ This assignment induces a map called the canonical evaluation map and through it, ${\displaystyle V}$ is identified as a vector subspace of the double dual ${\displaystyle V''.}$ The space ${\displaystyle V}$ is a semi-reflexive space if and only if this map is surjective. The ${\displaystyle f:X\to V}$ is Pettis integrable if and only if ${\displaystyle d_{A}\in V}$ for every ${\displaystyle A\in \Sigma .}$

An immediate consequence of the definition is that Pettis integrals are compatible with continuous linear operators: If ${\displaystyle \Phi \colon V_{1}\to V_{2}}$ is linear and continuous and ${\displaystyle f\colon X\to V_{1}}$ is Pettis integrable, then ${\displaystyle \Phi \circ f}$ is Pettis integrable as well and

$${\displaystyle \int _{X}\Phi (f(x))\,d\mu (x)=\Phi \left(\int _{X}f(x)\,d\mu (x)\right).}$$

The standard estimate

$${\displaystyle \left|\int _{X}f(x)\,d\mu (x)\right|\leq \int _{X}|f(x)|\,d\mu (x)}$$

for real- and complex-valued functions generalises to Pettis integrals in the following sense: For all continuous seminorms ${\displaystyle p\colon V\to \mathbb {R} }$ and all Pettis integrable ${\displaystyle f\colon X\to V}$,

$${\displaystyle p\left(\int _{X}f(x)\,d\mu (x)\right)\leq {\underline {\int _{X}}}p(f(x))\,d\mu (x)}$$

holds. The right-hand side is the lower Lebesgue integral of a ${\displaystyle [0,\infty ]}$-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\}.}

Taking a lower Lebesgue integral is necessary because the integrand ${\displaystyle p\circ f}$ may not be measurable. This follows from the Hahn-Banach theorem because for every vector ${\displaystyle v\in V}$ there must be a continuous functional ${\displaystyle \varphi \in V^{*}}$ such that ${\displaystyle \varphi (v)=p(v)}$ and for all ${\displaystyle w\in V}$, ${\displaystyle |\varphi (w)|\leq p(w)}$. Applying this to ${\displaystyle v:=\int _{X}f\,d\mu }$ 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 the convex hull of the values scaled by the measure of the integration domain:

$${\displaystyle \mu (A)<\infty {\text{ implies }}\int _{A}f\,d\mu \in \mu (A)\cdot {\overline {\operatorname {co} (f(A))}}}$$

This is a consequence of the Hahn-Banach theorem and generalizes the mean value theorem for integrals of real-valued functions: If ${\displaystyle V=\mathbb {R} }$, then closed convex sets are simply intervals and for ${\displaystyle f\colon X\to [a,b]}$, the following inequalities hold:

$${\displaystyle \mu (A)a~\leq ~\int _{A}f\,d\mu ~\leq ~\mu (A)b.}$$

Existence

If ${\displaystyle V=\mathbb {R} ^{n}}$ is finite-dimensional then ${\displaystyle f}$ is Pettis integrable if and only if each of ${\displaystyle f}$’s coordinates is Lebesgue integrable.

If ${\displaystyle f}$ is Pettis integrable and ${\displaystyle A\in \Sigma }$ is a measurable subset of ${\displaystyle X}$, then by definition ${\displaystyle f_{|A}\colon A\to V}$ and ${\displaystyle f\cdot 1_{A}\colon X\to V}$ are also Pettis integrable and

$${\displaystyle \int _{A}f_{|A}\,d\mu =\int _{X}f\cdot 1_{A}\,d\mu .}$$

If ${\displaystyle X}$ is a topological space, ${\displaystyle \Sigma ={\mathfrak {B}}_{X}}$ its Borel-${\displaystyle \sigma }$-algebra, ${\displaystyle \mu }$ a Borel measure that assigns finite values to compact subsets, ${\displaystyle V}$ is quasi-complete (that is, every bounded Cauchy net converges) and if ${\displaystyle f}$ is continuous with compact support, then ${\displaystyle f}$ is Pettis integrable. More generally: If ${\displaystyle f}$ is weakly measurable and there exists a compact, convex ${\displaystyle C\subseteq V}$ and a null set ${\displaystyle N\subseteq X}$ such that ${\displaystyle f(X\setminus N)\subseteq C}$, then ${\displaystyle f}$ is Pettis-integrable.

Let ${\displaystyle (\Omega ,{\mathcal {F}},\operatorname {P} )}$ 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 ${\displaystyle \operatorname {E} [v_{n}]}$ for the Pettis integral of ${\displaystyle v_{n}}$ (over ${\displaystyle X}$). Note that ${\displaystyle \operatorname {E} [v_{n}]}$ 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

$${\displaystyle \operatorname {E} [{\bar {v}}_{N}]={\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [v_{n}]\in V.}$$

Suppose that the partial sums

$${\displaystyle {\frac {1}{N}}\sum _{n=1}^{N}\operatorname {E} [{\bar {v}}_{n}]}$$

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 ${\displaystyle \langle \varphi ,\operatorname {E} [{\bar {v}}_{N}]-\lambda \rangle \to 0}$ for every functional ${\displaystyle \varphi \in V^{*}.}$ Consequently, ${\displaystyle \operatorname {E} [{\bar {v}}_{N}]\to \lambda }$ in the weak topology on ${\displaystyle X.}$

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

• Bochner measurable function
• Bochner integral
• Bochner space – Mathematical concept
• Vector measure
• Weakly measurable function

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• Sobolev, V. I. (2001) [1994], "Pettis integral", Encyclopedia of Mathematics, EMS Press