integral of spectral measure I have the following question: let $(X,\mathcal B,\mu)$ be a finite measure space and consider the operator $T_{\varphi} \colon L^2(X,\mu)\to L^2(X,\mu)$ given by $Tf(x)=\varphi(x)f(x)$, where $\varphi \in L^{\infty}(X,\mathcal{S},\mu)$. Consider the canonical spectral measure E induced by T. How we defined is as follows if $S \in B_{\sigma(T_{\varphi})}(T)$ then we define $E(S) = T_{1_{\varphi^{-1}}(S)}$ recall here $1_{\varphi^{-1}}(S)$ is the characteristic function.
I want to verify that $T_{\varphi} = \int_{\sigma(T)} z dE(z)$.
My attempt:
Suppose we have measurable partition $M_1,\ldots,M_n$ of $B_{\sigma(T_{\varphi})}(T)$ such that $|z_1 - z_2| < \epsilon$ for all $z_1,z_2 \in S_i$:
$$\|T_{\varphi} - \Sigma z_i E(M_i)\|$$
I am not sure why this is less than $\epsilon$. This is where I am stuck.

Note that $\sigma(T_\varphi)=\operatorname{ess ran}\varphi$. Given $x\in X$, there exists $j$ with  with $\varphi(x)\in M_j$. Then $|\varphi(x)-z_j|<\varepsilon$. Thus $x\in\varphi^{-1}(M_j)$ and then 
$$
|\varphi(x)-\sum_kz_k\, 1_{\varphi^{-1}(M_k)}(x)|=|\varphi(x)-z_j|<\varepsilon.
$$
So
\begin{align}
\|T_\varphi f-\sum_kz_k\, 1_{\varphi^{-1}(M_k)}\,f\|_2^2&=\int_{\sigma(T_\varphi)}|T_\varphi(x) f(x)-\sum_kz_k\, 1_{\varphi^{-1}(M_k)}(x)\,f(x)|^2\,d\mu\\[0.3cm]
&=\int_{\sigma(T_\varphi)}|T_\varphi(x) -\sum_kz_k\, 1_{\varphi^{-1}(M_k)}(x)|^2\,|f(x)|^2\,d\mu\\[0.3cm]
&\leq\varepsilon \|f\|_2^2.
\end{align}
As this works for all $f\in L^2$, we get that 
$$
\|T_\varphi -\sum_kz_k\, 1_{\varphi^{-1}(M_k)}\|<\varepsilon. 
$$
