Spectral Measures: Concentration Given a Hilbert space $\mathcal{H}$.
Consider spectral measures:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$
Define its support:
$$\operatorname{supp}(E):=\bigg(\bigcup_{U=\mathring{U}:E(U)=0}U\bigg)^\complement=\bigcap_{C=\overline{C}:E(C)=1}C$$
By second countability:
$$E\bigg(\operatorname{supp}E^\complement\bigg)\varphi=E\left(\bigcup_{k=1}^\infty B_k'\right)\varphi=\sum_{k=1}^\infty E(B_k')\varphi=0$$

But it may happen:
  $$\Omega\subsetneq\operatorname{supp}E:\quad E(\Omega)=E(\operatorname{supp}E)=1$$

What is an example?

Consider the operator $(Af)(x)=xf(x)$ on the Hilbert space $L^2([0,1])$ with Lebesgue measure. Then $E_A(M)=1_M$ is the spectral measure associated with $A$. Note that $\mbox{supp }E_A=\sigma(A)=[0,1]$, but $E_A(\{\lambda\})=0$ for any singleton $\lambda$, since $A$ does not have any eigenvalues. Now consider, say, the set $(0,1]$, for which $I=E([0,1])=E(\{0\})+E((0,1])=E((0,1])$.
