A question about spectral measure The following is a part of a theorem of Takesaki's Operator theory:
Let $T$ be an positive operator. Suppose $T = \int_0^{\|T\|} \lambda \, de(\lambda)$ is the spectral measure of $T$. Also put $e_\varepsilon = \int_\varepsilon^{\|T\|} d e(\lambda)$.  The arthur claims $$(T \xi,\xi) = \int_\varepsilon^{\infty} \lambda \, d\|e(\lambda)\xi\|^2 \geq \varepsilon \|\xi\|^2$$ for every $\xi\in e_\varepsilon H$, while I can not understand how the above equality holds and also why he calculates the integral until $\infty$ . Please help me. Thanks so much.
 A: The infinity in the integral is irrelevant, because $e$ is assumed to be supported on $[0,\|T\|]$. 
The equality uses the measure $\mu_\xi$ given by $\mu(E)=\langle e(E)\xi,\xi\rangle$. This is often used to even define what $e$ is. So, noting that 
$$
e_\epsilon T e_\epsilon=\int_0^{\|T\|}\lambda\,1_{[\epsilon,\|T\|]}(\lambda)\,de(\lambda)=\int_\epsilon^{\|T\|}\lambda\,\,de(\lambda)
$$
we get
$$
\langle T\xi,\xi\rangle=\langle Te_\epsilon\xi,e_\epsilon\xi\rangle=\langle e_\epsilon Te_\epsilon\,\xi,\xi\rangle
=\langle\left(\int_\epsilon^{\|T\|}\lambda\,\,de(\lambda)\right)\,\xi,\xi\rangle\\
=\int_\epsilon^{\|T\|}\lambda\,d\langle e(\lambda)\xi,\xi\rangle=\int_\epsilon^{\|T\|}\lambda\,d\|e(\lambda)\xi\|^2
$$
A: If $S$ is a Borel set, then
$$
              e(S)=e(S)^{2}=e(S)^{\star}.
$$
Therefore, $(e(S)x,x)=\|e(S)x\|^{2}$ for all $x$ because
$$
          \|e(S)x\|^{2} = (e(S)x,e(S)x)=(e(S)^{\star}e(S)x,x)=(e(S)^{2}x,x)=(e(S)x,x).
$$
You can approximate your integral $\int \lambda de(\lambda)$ with $\sum_{j}\lambda_{j}e(S_{j})$ and that leads to
$$
              \left(\int \lambda de(\lambda)\xi,\xi\right) \approx \sum_{j}\lambda_{j}\;(e(S_{j})x,x)=\sum_{j}\lambda_{j} \|e(S_{j})x\|^{2} \approx\int\lambda d\|e(\lambda)x\|^{2}.
$$
Taking limits gives
$$
       \left(\int \lambda de(\lambda)\xi,\xi\right) = \int\lambda d\|e(\lambda)x\|^{2}
$$
Because $\lambda \ge \epsilon$ on $[\epsilon,\|T\|]$,
$$
\begin{align}
      \left(\int_{\epsilon}^{\|T\|}\lambda de(\lambda)\xi,\xi\right)
       & = \int_{\epsilon}^{\|T\|}\lambda d\|e(\lambda)\xi\|^{2} \\
       & \ge \epsilon\int_{\epsilon}^{\|T\|}d\|e(\lambda)\xi\|^{2}
        = \epsilon\|e_{\epsilon}\xi\|^{2}.
\end{align}
$$
Therefore, if $e_{\epsilon}\xi = \xi$ (equivalently, $\xi \in e_{\epsilon}H$)
$$
           (T\xi,\xi) \ge \epsilon\|\xi\|^{2},\;\;\; \xi\in e_{\epsilon}H.
$$
