Direct integral of a measurable Hilbert field

Let $(X,\mathcal{B},\mu )$ carry a measurable complex Hilbert field $(H_x,e_n(x){)}_{x\in X}$ as in Measurable Hilbert field from a countable fundamental family, with inner products linear in the first variable. Write $\mathcal{M}(H)$ for its measurable sections. Define the square-integrable section space ${\mathcal{M}}_2(H):=\{\xi \in \mathcal{M}(H):{\int }_X\mathrm{\parallel }\xi (x){\mathrm{\parallel }}_{H_x}^{2}d\mu (x)<\mathrm{\infty }\}\mathrm{.}$ The integrand is a nonnegative measurable function: the section lemma Measurable sections have measurable pointwise inner products makes $x\mapsto \mathrm{\parallel }\xi (x)\mathrm{\parallel }$ measurable, and composition with $t\mapsto t^2$ preserves measurability by Composition with a Borel measurable outer map preserves measurability, whose measurability convention is inverse-image measurability A measurable function between measurable spaces. Its integral is the nonnegative Lebesgue integral The nonnegative Lebesgue integral.

For $\xi ,\eta \in \mathcal{M}(H)$, put $\xi {\sim }_{\mu }\eta$ when there is a Borel $\mu$-null set $N$ such that $\xi (x)=\eta (x)$ for every $x\mathrm{\notin }N$, using the noncomplete-base convention of the field definition. The direct integral is the quotient set

$${\int }_X^{\oplus }{H}_{x}d\mu (x):={\mathcal{M}}_2(H)\mathrm{/}{\sim }_{\mu }\mathrm{.}$$

Write $[\xi ]$ for the class of $\xi$. Addition and scalar multiplication are induced by pointwise fibre operations, and the inner product is $⟨[\xi ],[\eta ]⟩:={\int }_X⟨\xi (x),\eta (x){⟩}_{H_x}d\mu (x)\mathrm{.}$

The admissible sections form a vector space. Measurability of pointwise linear combinations follows from the section lemma. If $a=\mathrm{\parallel }\xi (x)\mathrm{\parallel }$ and $b=\mathrm{\parallel }\eta (x)\mathrm{\parallel }$, fibrewise expansion gives $\mathrm{\parallel }\xi (x)+\eta (x){\mathrm{\parallel }}^2=a^2+b^2+2\mathrm{Re}⟨\xi (x),\eta (x)⟩\le a^2+b^2+2ab\le 2(a^2+b^2)\mathrm{.}$ by Cauchy–Schwarz, $\mathrm{Re}z\le \mathrm{\mid }z\mathrm{\mid }$ for complex $z$, and $2ab\le a^2+b^2$ (since $(a-b{)}^2\ge 0$). The real-part bound follows from $\mathrm{\mid }u+iv\mathrm{\mid }=\sqrt{u^2+v^2}$ and $u^2+v^2\ge u^2$ when $u>0$; it is immediate when $u\le 0$, and taking nonnegative square roots gives $u\le \mathrm{\mid }u+iv\mathrm{\mid }$ in the positive case. The modulus and conjugation laws are recorded in Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive. By monotonicity, positive homogeneity, and additivity of the nonnegative integral (Monotonicity and nonnegative homogeneity of the nonnegative integral, Additivity of the nonnegative Lebesgue integral), the integral of the squared norm of a sum is finite whenever those of $\xi$ and $\eta$ are finite; the comparison function $a^2+b^2$ is measurable by Borel composition Composition with a Borel measurable outer map preserves measurability and measurable arithmetic Arithmetic and lattice operations preserve measurability whenever they are defined. For $c\mathrm{\ne }0$, $\mathrm{\parallel }c\xi (x){\mathrm{\parallel }}^2=\mathrm{\mid }c{\mathrm{\mid }}^2\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2$ and positive homogeneity keeps its integral finite; for $c=0$ the zero function has integral zero. Hence ${\mathcal{M}}_2(H)$ is a complex vector space. The pointwise bound uses Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs and the norm and scalar conventions of Real and complex inner-product spaces and their induced length.

The quotient and its operations are well-defined. The empty set is a Borel null set by the $m=0$ clause of Finite and countable subadditivity of measures, so ${\sim }_{\mu }$ is reflexive; it is symmetric by equality, and it is transitive because two Borel null witnesses have a null union by the same finite-subadditivity theorem. This uses the definition of almost-everywhere equality Measure-null sets and almost-everywhere statements relative to a measure. If $\xi {\sim }_{\mu }{\xi }^{\mathrm{\prime }}$ and $\eta {\sim }_{\mu }{\eta }^{\mathrm{\prime }}$, then outside the union of their two witnesses the pointwise sums agree, and so do the pointwise scalar multiples. The same finite union argument shows that these operations do not depend on representatives.

The displayed pairing is finite. By the section lemma, $h(x)=⟨\xi (x),\eta (x)⟩$ is measurable. The modulus $\mathrm{\mid }h\mathrm{\mid }$ is measurable by Borel composition with the complex modulus. The measurable real-valued functions $f(x)=\mathrm{\parallel }\xi (x)\mathrm{\parallel }$ and $g(x)=\mathrm{\parallel }\eta (x)\mathrm{\parallel }$ belong to ${\mathcal{L}}^2(\mu )$ by the definition of that space The function space ${\mathcal{L}}^p(\mu )$ for $0<p<\mathrm{\infty }$. Fibrewise Cauchy–Schwarz and then the scalar $L^2$ Cauchy–Schwarz inequality Cauchy-Schwarz inequality for $L^2$ give ${\int }_X\mathrm{\mid }h(x)\mathrm{\mid }d\mu (x)\le {\int }_{X}f(x)g(x)d\mu (x)\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2<\mathrm{\infty }\mathrm{.}$ Since $f,g\ge 0$, the scalar theorem's $\mathrm{\mid }fg\mathrm{\mid }$ is $fg$. Here $fg$ is measurable by Arithmetic and lattice operations preserve measurability whenever they are defined, and the first inequality uses monotonicity of the nonnegative integral Monotonicity and nonnegative homogeneity of the nonnegative integral. Thus $h$ is an integrable complex function by Integrable real and complex functions, and their integrals. More explicitly, write $h=u+iv$. The real coordinate projections are Borel, so $u$ and $v$ are measurable, and $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }$ makes both real functions integrable. Their classes, rather than a class of the complex function $h$, belong to the real space $L^1(\mu )$ of The space $L^p(\mu )$ as the quotient by null functions. The complex integral here means precisely $\int h=\int u+i\int v$.

The pairing does not depend on representatives. If $\xi {\sim }_{\mu }{\xi }^{\mathrm{\prime }}$ and $\eta {\sim }_{\mu }{\eta }^{\mathrm{\prime }}$, their pairings agree outside the union of the two null witnesses. Both pairings are integrable by the preceding estimate. Their real parts agree almost everywhere, as do their imaginary parts. Applying Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree separately to these real $L^1$ classes, with $A=X$, gives equality of both component integrals and hence of the complex integrals.

The quotient is an inner-product space. Fibrewise linearity in the first variable passes through the integral by real linearity in The Lebesgue integral is linear on $L^1(\mu )$, applied to the integrable real and imaginary parts. Indeed, if $h=u+iv$ and $c=a+ib$, then $ch=(au-bv)+i(av+bu)$, so real linearity gives $\int ch=c\int h$. For two integrable pairings $h=u+iv$ and $k=s+it$, $h+k=(u+s)+i(v+t)$ gives $\int (h+k)=\int h+\int k$ in the same way; all these real components are integrable by real linearity. Their complex combinations are integrable since $\mathrm{\mid }ch\mathrm{\mid }=\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }h\mathrm{\mid }$ and $\mathrm{\mid }h+k\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }k\mathrm{\mid }$, using nonnegative integral monotonicity and additivity. Fibrewise conjugate symmetry passes through it as well: for integrable $h=u+iv$, the definition gives $\int \bar{h}=\int u-i\int v=\overline{\int h}$; integrability of $\bar{h}$ follows from $\mathrm{\mid }\bar{h}\mathrm{\mid }=\mathrm{\mid }h\mathrm{\mid }$ Real and imaginary parts, complex conjugation, and modulus and Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive. Conjugate symmetry and first-variable linearity give conjugate-linearity in the second variable. For every class, $⟨[\xi ],[\xi ]⟩={\int }_X\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^{2}d\mu (x)\ge 0.$ This value is zero exactly when $\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2=0$ almost everywhere, by A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. Fibrewise positive definiteness says this is exactly when $\xi (x)=0$ off a Borel null set, that is, when $[\xi ]=[0]$. Therefore the pairing is positive definite. These arguments use the ordinary complex inner-product axioms Real and complex inner-product spaces and their induced length and the integral convention above; they do not use any form of the axiom of choice.

No completion is part of this definition. The next theorem proves that this inner-product space is complete under its induced norm.

Boundary cases. If $X=\mathrm{\varnothing }$, or if every fibre is zero-dimensional, there is only the zero section and the quotient is the zero inner-product space. If $\mu (X)=0$, then $X$ itself is a Borel null set, so every two square-integrable sections are equivalent and the quotient is again zero. For a one-point base $X=\{x_0\}$ with $\mu (\{x_0\})=m>0$ and $H_{x_0}=\mathbb{C}$, take $e_1(x_0)=1$ and $e_n(x_0)=0$ for $n\ge 2$. Every section has the form ${\xi }_z(x_0)=z$ and $⟨[{\xi }_z],[{\xi }_w]⟩=mz\bar{w}\mathrm{.}$ The direct integral is one-dimensional, with norm $\sqrt{m}\mathrm{\mid }z\mathrm{\mid }$. If instead $m=0$, the preceding zero-measure calculation applies.