Given a Borel space $\Omega$.
Regard the Hilbert Space: $$\mu:\mathcal{B}(\Omega)\to\overline{\mathbb{R}}_+:\quad\mathcal{H}:=\mathcal{L}^2(\Omega;\mu)$$
Denote the Borel Projections: $$E:\mathcal{B}(\Omega)\to\mathcal{B}(\mathcal{H}):\quad E(A)\varphi:=1_A\varphi$$
Then for Scalar Operators: $$T\in\mathcal{B}(\mathcal{H}):\quad T=M_\tau\iff ET=TE$$ How can I prove this equivalence?
If $\mu$ is sigma-finite, then there are disjoint Borel subsets $\{ A_j \}_{j=1}^{\infty}$ of finite $\mu$-measure such that $\bigcup_j A_j=\Omega$. Let $f_j = T1_{A_j}$. Then $$ 1_{A_k}f_j = 1_{A_k}T1_{A_j}=T1_{A_k\cap A_j} =0 ,\;\;\; k \ne j,\\ 1_{A_j}f_j = f_j. $$ So each function $f_j$ is supported in $A_j$. Let $f$ be the a.e. unique Borel function for which $1_{A_j}f=f_j$. If $A$ is a Borel subset of finite measure, then $1_A=\sum_{j=1}^{\infty}1_A1_{A_j}$ is an orthogonal expansion in $L^2(\Omega,\mu)$. Because $T$ is continuous on $L^2(\Omega,\mu)$, $$ T 1_A = T\sum_{j=1}^{\infty}1_{A_j}1_{A}=\sum_{j=1}^{\infty}T(1_{A_j}1_{A})=\sum_{j=1}^{\infty}1_Af_j=f1_A. $$ Therefore $T\varphi = f\varphi$ for any simple function $\varphi = \sum_{k=1}^{K}\alpha_k\chi_{E_k}$ where $\mu E_k < \infty$ for all $k$.