Integral of a simple function against a pvm

Assume Countable Choice. Let $(X,\mathrm{\Sigma })$ be a measurable space, let $H$ be a complex Hilbert space, let $E$ be a projection valued measure on $(X,\mathrm{\Sigma })$, and let $s:X\to \mathbb{C}$ be a complex simple function (Complex simple functions as finite sums of measurable indicators, Measurable spaces and measurable sets). Present $s$ in disjoint normal form with a zero-coefficient complement, that is

$s={\sum }_{j=1}^m{a}_j{\mathbf{1}}_{B_j},$

where $m\ge 1$, and $B_1,\dots ,B_m\in \mathrm{\Sigma }$ are pairwise disjoint with $B_1\cup \cdots \cup B_m=X$ and $a_1,\dots ,a_m\in \mathbb{C}$; a representation over a disjoint family whose union misses some measurable set is completed by adding that set with the coefficient $0$, and any coefficient is allowed to vanish. In particular the zero function on the empty space uses $m=1$, $B_1=\mathrm{\varnothing }$ and $a_1=0$; an empty presentation is not used. Then define

$\int sdE:={\sum }_{j=1}^m{a}_{j}E(B_j)\in \mathcal{B}(H)\mathrm{.}$

Here ${\mathbf{1}}_B$ is the indicator of $B$ and $E(B_j)$ is the value of the projection valued measure on $B_j$ (Projection valued measure).

Well-definedness. Because the $B_j$ are pairwise disjoint and cover $X$, the operator ${\sum }_j{a}_{j}E(B_j)$ is a finite sum of bounded operators and hence a bounded operator; the sum is meaningful in $\mathcal{B}(H)$ with the operator norm, and $\mathrm{\parallel }{\sum }_j{a}_{j}E(B_j)\mathrm{\parallel }\le {\max }_j\mathrm{\mid }{a}_j\mathrm{\mid }$, since the projection values are contractive and pairwise orthogonal. The value displayed a priori depends on the chosen disjoint presentation of $s$; that it does not, $\int sdE$ is independent of the disjoint presentation of $s$, is proved as Simple pvm integral is representation independent immediately below, before the symbol is used. The normal-form convention $s={\sum }_j{a}_j{\mathbf{1}}_{B_j}$ with $B_1\cup \cdots \cup B_m=X$ is the one used throughout this page, and the identity $\int {\mathbf{1}}_{B}dE=E(B)$ holds for every $B\in \mathrm{\Sigma }$.