> [!definition] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]], $(\Omega, \cm)$ be a [[Sigma Algebra|measurable space]]. A mapping $P: \cm \to L(\ch, \ch)$ is a $\ch$-**projection-valued measure** if: > 1. Each $P(E)$ is an [[Orthogonal Projection|orthogonal projection]]. > 2. $P(\emptyset) = 0$ and $P(\Sigma) = 1$. > 3. $P(E \cap F) = P(E)P(F)$. > 4. If $\seq{E_j} \subset \cm$ are disjoint, then $P\paren{\bigcup_{n \in \nat}E_n} = \sum_{n \in \nat}P(E_n)$, where the sum converges in the [[Operator Topologies|strong operator topology]] unconditionally. > [!theorem] > > If $E, F \in \cm$ are disjoint, then $P(E)$ and $P(F)$ have orthogonal ranges. > > *Proof*. Let $x, y \in \ch$, then > $ > \angles{P(E)x, P(F)y} = \angles{P(E)P(F)x, y} = \angles{P(\emptyset)x, y} = 0 > $ > [!definition] > > Let $f: \Omega \to \complex$ be a [[Simple Harmonic Motion|simple]] Borel function with $f = \sum_{j = 1}^n a_j \one_{E_j}$, define the **integral** of $f$ by > $ > \angles{f, P} = \sum_{j = 1}^n a_j P(E_j) > $ > Then there exists a unique extension of the integral to $B(\Omega)$ as a [[Bounded Linear Map|bounded linear map]] with respect to the [[Uniform Convergence|uniform]] norm on $B(\Omega)$ and the [[Space of Bounded Linear Maps|operator norm]] on $L(\ch, \ch)$. > > *Proof*. For each $x \in \ch$, this defines a linear map $x \mapsto \angles{f, P}(x)$. Assuming that $\seqf{E_j}$ is a disjoint family, then by Parseval's identity, > $ > \begin{align*} > \angles{f, P}(x) &= \sum_{j = 1}^n a_jP(E_j)(x) \\ > \norm{\angles{f, P}(x)}^2&\le \norm{x}^2 \cdot \max_{1 \le j \le n}\abs{a_j}^2 > \end{align*} > $ > after completing to a Hilbert basis. By the [[Linear Extension Theorem|linear extension theorem]], there exists a unique extension of $\angles{\cdot, P}(x)$ to the uniform closure of the simple functions, which is $B(\Omega)$. > > For each $f \in B(\Omega)$, this defines a new map that is a pointwise limit of linear maps, hence it is also linear. Given that the operator norm of the image is bounded by the uniform norm of the function, uniform convergence of the functions implies that the resulting operator is also bounded. > [!definition] > > Let $x, y \in \ch$, define > $ > P_{x, y}(E) = \angles{P(E)x, y} > $ > then $P_{x, y}$ is a [[Complex Measure|complex measure]] by $(4)$. $P$ is **regular** if $P_{x, y}$ is regular for all $x, y \in \ch$. > > The mapping $(x, y) \mapsto P_{x, y}$ is a sesquilinear map such that $P_{x, x}$ is positive. Moreover, for any $f \in B(\Sigma)$, > $ > \angles{f, P_{x, y}} = \angles{\angles{f, P}x, y} > $ > and for any uniformly bounded sequence $\seq{f_n} \in B(\Sigma)$ such that $f_n \to f$ pointwise, $\angles{f_n, P} \to \angles{f, P}$ in the weak operator topology. > > *Proof*. Firstly, for any simple function $f$, $\angles{f, P_{x,y}} = \angles{\angles{f, P}x, y}$. If $f$ is an arbitrary bounded Borel function, then there exists a sequence of simple functions $\seq{f_n}$ such that $f_n \to f$ uniformly, in which case > $ > \angles{f_n, P_{x, y}} \to \angles{f, P_{x, y}} \quad \angles{f_n, P} \to \angles{f, P} > $ > so $\angles{f, P_{x, y}} = \angles{\angles{f, P}x, y}$. > [!theorem] > > The mapping $f \mapsto \angles{f, P}$ is a $*$-homomorphism from $B(\Omega)$ to $L(\ch)$. > > *Proof*. Let $f = \sum_{i = 1}^m a_i\one_{E_i}$ and $g = \sum_{j = 1}^n b_j\one_{F_j}$ be simple functions, where $\bracs{E_i}, \bracs{F_j}$ are assumed to be disjoint families. In this case, > $ > \begin{align*} > \angles{fg, P} &= \sum_{i, j}a_ib_jP(E_i \cap F_j) = \sum_{i, j}a_ib_jP(E_i)P(F_j) \\ > &= \angles{f, P}\angles{g, P} > \end{align*} > $ > Passing through uniform limits yields the result. Similarly, $\angles{\ol f, P} = \angles{f, P}^*$. > [!theorem] > > Let $\seq{f_n} \subset B(\Omega)$. If $\seq{f_n}$ is uniformly bounded and converges pointwise to $f$, then $\angles{f_n, P} \to \angles{f, P}$ in the strong operator topology. > > *Proof*. $\angles{f_n, P} = \Gamma^{-1}f \to \Gamma f = \angles{f, P}$ in the strong operator topology.