A <projection-valued measure> on the <Borel sets> of $\mathbb C$ is a map $E$ into the <orthogonal projections> on a <separable Hilbert space> $\mathcal H$ such that
$$
E(\varnothing)=0,\qquad E(\mathbb C)=I,\qquad
E(B\cap C)=E(B)E(C),
$$
and for pairwise disjoint $B_j$,
$$
E\!\left(\bigcup_jB_j\right)v=\sum_jE(B_j)v
$$
for every $v\in\mathcal H$, with convergence in norm.

The <spectral theorem for normal operators on a separable Hilbert space> states that a bounded <normal operator> $A$ has a unique projection-valued measure supported on $\operatorname{Sp}(A)$ for which
$$
\boxed{A=\int_{\operatorname{Sp}(A)}z\,dE(z).}
$$
More generally, the <Borel functional calculus for a normal operator> is
$$
f(A)=\int f(z)\,dE(z).
$$

For $v,w\in\mathcal H$, the <scalar spectral measures> are
$$
\mu_{v,w}(B)=\langle E(B)v,w\rangle,
\qquad
\mu_v=\mu_{v,v}.
$$
If $A$ is <self-adjoint operator>[self-adjoint], its spectrum and hence the support of $E$ lie in $\mathbb R$. Moreover,
$$
\mu_v(B)=\langle E(B)v,v\rangle=\|E(B)v\|^2\geq0,
$$
so $\mu_v$ is a <positive measure>, and
$$
\boxed{\mu_v(\mathbb R)=\langle Iv,v\rangle=\|v\|^2.}
$$
The paper prints total mass $\|v\|$; with the standard definition it is $\|v\|^2$, so the unsquared norm is a typographical error.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 358 / 2 / a / Solution

A Hilbert space is separable when it has a countable dense subset, equivalently a countable <Hilbertian basis>.

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 Incoming links: Separable Hilbert space