Bounded borel pvm integral

Uniform simple approximation: for each $n\ge 0$ choose finitely many pairwise disjoint measurable sets $D_1,\dots ,D_r$ covering $X$ and complex numbers $c_i$ with $\mathrm{\mid }f-c_i\mathrm{\mid }\le 1\mathrm{/}(n+1)$ on $D_i$ (cut a square containing the range of $f$ into finitely many squares of diameter $<1\mathrm{/}(n+1)$ and take inverse images), so $s_n:={\sum }_i{c}_i{\mathbf{1}}_{D_i}$ is a complex simple function with $\mathrm{\parallel }f-s_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}(n+1)$.

Difference of simple integrals: if $s,t$ are complex simple functions, presenting both over the common refinement of their disjoint normal forms gives $\int (s-t)dE=\int sdE-\int tdE$, hence $\mathrm{\parallel }\int sdE-\int tdE\mathrm{\parallel }\le \mathrm{\parallel }s-t{\mathrm{\parallel }}_{\mathrm{\infty }}$; in particular the sequence $\int s_{n}dE$ is Cauchy, since $\mathrm{\parallel }\int s_{n}dE-\int s_{m}dE\mathrm{\parallel }\le \mathrm{\parallel }{s}_n-s_m{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{n+1}+\frac{1}{m+1}$.

By completeness of $\mathcal{B}(H)$ the sequence $\int s_{n}dE$ has a norm limit ${\mathrm{\Phi }}_E(f)$, and for any other uniformly approximating sequence $(t_n)$ one has $\mathrm{\parallel }\int s_{n}dE-\int t_{n}dE\mathrm{\parallel }\le \mathrm{\parallel }{s}_n-t_n{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so the limit does not depend on the sequence; the same argument applies to the difference of two candidate limits.

Pairing identity: for all $x,y\in H$, $⟨{\mathrm{\Phi }}_E(f)x,y⟩={\lim }_n⟨(\int s_{n}dE)x,y⟩={\lim }_n\int s_{n}d{E}_{x,y}=\int fd{E}_{x,y}$, because $\mathrm{\mid }\int (s_n-f)d{E}_{x,y}\mathrm{\mid }\le \mathrm{\parallel }{s}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{E}_{x,y}\mathrm{\mid }(X)\le \frac{1}{n+1}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }\to 0$.

Norm identities: $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)x{\mathrm{\parallel }}^2={\lim }_n\mathrm{\parallel }(\int s_{n}dE)x{\mathrm{\parallel }}^2={\lim }_n\int \mathrm{\mid }{s}_n{\mathrm{\mid }}^{2}d{E}_x=\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d{E}_x$ by dominated convergence applied to the finite measure $E_x$ with the constant dominating function $(M+1{)}^2$, since $s_n\to f$ and $\mathrm{\mid }{s}_n\mathrm{\mid }\le M+1$; taking $y=x$ in the pairing identity gives $⟨{\mathrm{\Phi }}_E(f)x,x⟩=\int fd{E}_x$.

Norm bound and uniqueness: $\mathrm{\mid }⟨{\mathrm{\Phi }}_E(f)x,y⟩\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by the pairing identity just proved and the variation bound, so $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$; and any operator $T$ with $⟨Tx,y⟩=\int fd{E}_{x,y}$ for all $x,y$ equals ${\mathrm{\Phi }}_E(f)$, since $T-{\mathrm{\Phi }}_E(f)$ has all pairings zero, whence $(T-{\mathrm{\Phi }}_E(f))x=0$ for every $x$ by positive definiteness of the pairing.

Upper bound for the norm: for $x\mathrm{\ne }0$ one has $E_x=\mathrm{\parallel }x{\mathrm{\parallel }}^2{E}_{x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }}$ and hence $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },E_x}=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },E_{x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }}}$, so $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)x{\mathrm{\parallel }}^2=\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d{E}_x\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },E_x}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_{E,\mathrm{\infty }}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by the least-essential-bound property; therefore $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{E,\mathrm{\infty }}$.

Lower bound for the norm: write $a=\mathrm{\parallel }f{\mathrm{\parallel }}_{E,\mathrm{\infty }}$. If $a=0$, the lower bound follows from nonnegativity of the operator norm. If $a>0$, given $0\le c<a$ choose a unit vector $x$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },E_x}>c$, so $A:=\{\mathrm{\mid }f\mathrm{\mid }>c\}$ satisfies $E_x(A)>0$ by the least-essential-bound property; put $y:=E(A)x\mathrm{\ne }0$, so $E_y(X\setminus A)=0$ and hence $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)y{\mathrm{\parallel }}^2=\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d{E}_y\ge c^2{E}_y(X)=c^2\mathrm{\parallel }y{\mathrm{\parallel }}^2$, giving $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)\mathrm{\parallel }\ge c$; since $0\le c<a$ was arbitrary, $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)\mathrm{\parallel }\ge \mathrm{\parallel }f{\mathrm{\parallel }}_{E,\mathrm{\infty }}$.

The integral ${\mathrm{\Phi }}_E(f)$ is well defined, obtained from uniform simple approximations, satisfies the pairing and quadratic identities and the bound $\mathrm{\parallel }{\mathrm{\Phi }}_E(f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and its exact norm is the $E$-essential supremum $\mathrm{\parallel }f{\mathrm{\parallel }}_{E,\mathrm{\infty }}$.