= Spectral theorem for normal operators on a separable Hilbert space
A <normal operator> $A$ on a separable complex <Hilbert space> has a unique normalized <projection-valued measure> $E$ such that
$$
A=\int_{\mathbb C}z\,dE(z).
$$
For a bounded normal operator, $E$ is supported on its compact <spectrum> and the integral defines an everywhere-defined bounded operator. For a closed densely defined normal operator, its <operator domain> is
$$
D(A)=\left\{x:\int_{\mathbb C}|z|^2\,d\langle E(z)x,x\rangle<\infty\right\}.
$$
The scalar measure in this expression is positive and has total mass $\|x\|^2$.
A spectral integral can exist without a single eigenvector. On $L^2(0,1)$, $(Af)(t)=tf(t)$ is bounded and self-adjoint, and $E(B)f=\mathbf1_B f$. An eigenvector for $\lambda$ would have to be supported on $\{t=\lambda\}$, a set of Lebesgue measure zero, so it would vanish in the <L2 space>. The spectral measure therefore extends orthonormal diagonalization to continuous spectral parameters rather than claiming an eigenbasis always exists.
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