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?
 A: 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 $I=E([0,1])=E(\{0\})+E((0,1])=E((0,1])$.
A: I'm assuming that your spectral measure is defined on Borel subsets of $\mathbb{C}$, which is the standard definition of the spectral measure which is the spectral resolution for a normal operator $T$.
You have defined the support $\mbox{supp}E$ to be the intersection of all closed subsets $C$ for which $E(C)=I$. If $E(A)=I$ for some closed subset $A\subsetneq \mbox{supp}E$, then $A$ is one of the sets in your intersection, yielding an obvious contradiction.
If $\mbox{supp}E$ is contained in some closed disk of finite radius $R$, then $T_{E}=\int \lambda dE(\lambda)$ satisfies
$$
\begin{align}
        \|T_{E}x\|^{2} & = \int_{\mbox{supp}E} |\lambda|^{2}d\|E(\lambda)x\|^{2} \\
              & \le \int_{\mbox{supp}E}d\|E(\lambda)x\|^{2}R^{2} = \|E(\mbox{supp}E)x\|^{2}R^{2} = R^{2}\|x\|^{2}.
\end{align}
$$
So $\|T_{E}\| \le \|R\|$.
