Given a Hilbert space $\mathcal{H}$ and spectral a measure $E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$.
Define the integral of simple functions by: $$\int_\Omega s\mathrm{d}E:=\sum_{z\in\mathbb{C}}z\cdot E(s^{-1}z)\quad(s:\Omega\to\mathbb{C})$$ and extend it to the uniform closure: $$\int_\Omega f\mathrm{d}E:=\lim_{\lambda\in\Lambda}\int_\Omega s_\lambda\mathrm{d}E\quad(s_\lambda\stackrel{\infty}{\to}f)$$ This goes fine as: $$\|\int_\Omega s\mathrm{d}Ex\|^2=\ldots\leq\|s\|_\infty^2\cdot\sum\|E(A_k)x\|^2=\ldots\leq\|s\|_\infty^2\cdot\|x\|^2$$
How to prove then that it respects multiplication: $$\int_\Omega fg\mathrm{d}E=\int_\Omega f\mathrm{d}E\cdot\int_\Omega g\mathrm{d}E$$
The problem I'm having is that I hardly even can imagine this to hold for the Lebesgue integral: $$\int_\Omega fg\mathrm{d}µ=\int_\Omega f\mathrm{d}µ\cdot\int_\Omega f\mathrm{d}µ$$
Hint: suppose first that $f,g$ are simple functions. Verify that $f,g$ can be written as $$f = \sum_{i=1}^n a_i 1_{A_i}, \qquad g = \sum_{i=1}^n b_i 1_{A_i}$$ for some common collection of measurable sets $\{A_1, \dots, A_n\}$, which moreover can be taken to be pairwise disjoint.
Then expand the product in $\int fg\,dE$ and recall the property $E(A) E(B) = 0$ when $A,B$ are disjoint. That will get rid of the cross terms.