Integral of a measurable function against a projection-valued measure

Assume Countable Choice. Let $E$ be a projection valued measure on the measurable space $(X,\mathrm{\Sigma })$ acting on the complex Hilbert space $H$ (Projection valued measure), and let $f:X\to \mathbb{C}$ be $\mathrm{\Sigma }$-measurable. Put $D(f(E)):=\{x\in H:{\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{2}d{E}_x<\mathrm{\infty }\},E_x(B)=⟨E(B)x,x⟩,$ where the scalar measures $E_x$ are those of Scalar and complex measures from a pvm, and for $x\in D(f(E))$ set $f(E)x:={\lim }_{n\to \mathrm{\infty }}{\mathrm{\Phi }}_E\left(f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le n\}}\right)x,$ For $H\mathrm{\ne }\{0\}$, the bounded integrals ${\mathrm{\Phi }}_E$ are those of Bounded borel pvm integral, and the limit is taken in the norm of $H$. For $H=\{0\}$, define ${\mathrm{\Phi }}_E(g)$ to be the unique operator on $H$ for every bounded measurable $g$. Then $E_0$ is the zero measure, $D(f(E))=H$, and $f(E)0=0$. Thus this case is defined directly without applying a theorem requiring a nonzero space.

Well-definedness. $D(f(E))$ is a linear subspace: the estimate $E_{x+y}\le 2E_x+2E_y$ for scalar measures and $E_{\alpha x}=\mathrm{\mid }\alpha {\mathrm{\mid }}^2{E}_x$ follow from $\mathrm{\parallel }E(B)(x+y){\mathrm{\parallel }}^2\le 2\mathrm{\parallel }E(B)x{\mathrm{\parallel }}^2+2\mathrm{\parallel }E(B)y{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }E(B)\alpha x{\mathrm{\parallel }}^2=\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\parallel }E(B)x{\mathrm{\parallel }}^2$. Writing $f_n:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le n\}}$ the sets $\{\mathrm{\mid }f\mathrm{\mid }\le n\}$ are measurable, so $f_n$ is bounded and measurable. For $m,n\ge N$, $\mathrm{\mid }{f}_n-f_m{\mathrm{\mid }}^2\le \mathrm{\mid }f{\mathrm{\mid }}^2{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>N\}}\mathrm{.}$ The integral of the right side against $E_x$ tends to zero by Dominated convergence, with the integrable majorant $\mathrm{\mid }f{\mathrm{\mid }}^2$ and pointwise limit zero since $f$ is finite-valued. Linearity of the bounded calculus (Pvm integral is a star homomorphism) and its quadratic identity give $\mathrm{\parallel }{\mathrm{\Phi }}_E(f_n)x-{\mathrm{\Phi }}_E(f_m)x{\mathrm{\parallel }}^2=\int \mathrm{\mid }{f}_n-f_m{\mathrm{\mid }}^{2}d{E}_x\mathrm{.}$ These clauses hold directly on the zero space as well. Hence the truncation vectors are Cauchy and converge uniquely by Hilbert-space completeness (Hilbert space). The limit is linear in $x$ because every truncation operator is linear and addition and scalar multiplication are norm-continuous. Countable Choice is inherited from the bounded PVM suppliers; no further choice is needed to take these specified limits.

Finally, if two $\mathrm{\Sigma }$-measurable functions $f,g$ agree outside a measurable set $N$ with $E(N)=0$, then $E_x(N)=\mathrm{\parallel }E(N)x{\mathrm{\parallel }}^2=0$ for every $x$. Thus the integrals of $\mathrm{\mid }f{\mathrm{\mid }}^2$ and $\mathrm{\mid }g{\mathrm{\mid }}^2$ agree, so their domains agree. At every truncation level their bounded truncations agree outside $N$; the quadratic identity applied to the difference gives equal truncation vectors for every $x$. Taking limits gives $f(E)x=g(E)x$ on their common domain.