# 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 $$\|T_\varphi -\sum_kz_k\, 1_{\varphi^{-1}(M_k)}\|<\varepsilon.$$