I am reading about the spectral theory of bounded linear operators in a book called Quantum Theory for Mathematicians by Hall. The goal of the spectral theory presented by the author is to construct a reasonable notion for what it means to "pass a bounded/unbounded" operator to the argument of some function. Namely, if $A\in B(H)$ is a bounded linear operator and $f$ is, say, a continuous complex-valued function, then
$$f(A) = \int_{\sigma(A)}f(\lambda)d\mu^A(\lambda)$$
for a unique projection valued measure $\mu^A$ which satisfies $A = \int_{\sigma(A)}\lambda d\mu^A(\lambda)$ with $\sigma(A)$ being the spectrum of $A$. Theory and related proofs for the case of bounded linear operators is given in chapters 7 and 8.
For the rest of this question I am assuming what you know how projection valued measures and the related tools work (since otherwise I would probably have to copy most of the book here). What is puzzling me is how the integral $\int_{\sigma(A)}f(\lambda)d\mu^A(\lambda)$ of a non-simple function, complex-valued, function is defined. I mean the following: If $f = \sum_{n=0}^N\alpha_n\chi_{E_n}$ for some measurable subsets of $\sigma(A)$ $E_1,\dots,E_N$ and $\alpha_1,\dots,\alpha_N\in\mathbb{C}$, then
$$\int_{\sigma(A)}f(\lambda)d\mu^A(\lambda) = \sum_{n=0}^\infty \alpha_n\mu^A(E_n)$$
which is a bounded linear operator as the finite sum of such operators. The definition which I have seen for the Lebesgue integral of a non-negative measurable function $f:X\to[0,\infty]$ in a measure space $(X,\mathcal{F},\mu)$ is
$$\int_Xf(x)d\mu(x) = \sup\left\{\int_Xs(x)d\mu(x)\mid \text{$s:X\to[0,\infty]$ is a simple function and $\forall x\in X:0\leq s(x)\leq f(x)$}\right\}$$
And leads to this post: Since Hall does not define from which parts the integral of non-simple function is constructed from, I am lead to believe that we are sameish supremum "under the hood". But w.r.t. what should the supremum be taken against?