Let $(X, \mathcal A, E, \mathcal H)$ be a spectral measure space. Let $\varphi\in L^{\infty} (E)$ be a complex valued bounded $\mathcal A$-measurable function on $X.$ Let $J_{\varphi} : = \int_X \varphi\ dE.$ Then show that $\|J_{\varphi}\| = \text {E-}\sup |\varphi|.$
What I know is that for any complex valued $\mathcal A$-measurable function $\varphi$ on $X$ we have $$\left \|J_{\varphi} x \right \|^2 = \int_{X} |\varphi|^2\ d\mu_x$$
where $\mu_x : \mathcal A \longrightarrow [0,\infty]$ is a measure defined by $$\mu_x (A) : = \left \langle x, E(A) x \right \rangle = \left \|E(A) x \right \|^2, \ x \in \mathcal H, A \in \mathcal A.$$
From here we can say that $$\|J_{\varphi}\| \leq \text {E-}\sup |\varphi|.$$
But I find difficulty in proving the equality? Could anybody give me some suggestion regarding that?
Thanks in advance.
EDIT $:$ After some work I can able to do it for simple $\mathcal A$-measurable functions. Let $\varphi = \sum\limits_{i=1}^{n} c_i \chi_{A_i}$ be a bounded simple $\mathcal A$-measurable function, where $c_i \in \mathbb C$ and $A_i \in \mathcal A$ with $A_i \cap A_j = \varnothing,$ for $i \neq j.$ If $\text {E-}\sup |\varphi| = 0$ then we have nothing to prove because then both sides are trivially zero. So let us assume that $\text {E-}\sup |\varphi| \neq 0.$ WLOG we may assume that $\text {E-}\sup |\varphi| = |c_1|.$ Then $E(A_1) \neq 0.$ Let $x \in \left (\text {Ker}\ (E(A_1)) \right )^{\perp}.$ Then we have $E(A_1) x = x$ and $E(A_j) x = 0,$ for $j \neq 1.$ So we have $$\|J_{\varphi} x\| = |c_1| \|x\|.$$ This shows that $$\|J_{\varphi}\| = |c_1| = \text {E-}\sup |\varphi|.$$
But I got stuck in generalizing this result for any non-negative $\mathcal A$-measurable function by using simple function technique. For a general non-negative $\mathcal A$-measurable function $\varphi$ if we try to generalize it by an increasing sequence of non-negative simple $\mathcal A$-measurable functions $\left \{\varphi_n \right \}_{n \geq 1}$ then I find that $$\|J_{\varphi}\| = \lim\limits_{n \to \infty} \text {E-}\sup |\varphi_n|.$$ But can we say the following $$\lim\limits_{n \to \infty} \text {E-}\sup |\varphi_n| = \text {E-}\sup |\varphi|.$$