Let $(X,\mathscr{M},\mu)$ be a measure space. When defining the Lebesgue integral, one first defines it for class $\mathscr{S}_+$ of non negative (real-valued) simple functions and move on to defining it for $[0,\infty]$-valued functions (all functions are measurable unless otherwise stated). The definition is of course $$\int f \, d\mu=\sup\left\{\int s \, d\mu \,|\,s \leq f, \ s\in\mathscr{S}_+ \right\}.$$
Now let $f,g:X\rightarrow [0,\infty]$. The additivity $\int f+g \, d\mu=\int f \, d\mu + \int g \, d\mu$ is of course valid, but all the proofs that I'm aware of use the (simple version of) monotone convergence theorem to establish this. Is there a way to avoid such uses of convergence theorems?; perhaps by defining upper and lower integrals independently and declaring a function to be integrable iff two integrals coincide?
I'm perfectly happy with the definition of Lebesgue integral, but the reason I'm asking this question is that I've faced a situation where taking limits are not allowed. To explain my situation:
Let $X$ be a set and $\mu$ be a FINITELY additive function on $\mathscr{P}(X)$, all subsets of $X$, such that $\mu(X)=1$ (I did not use the word 'measure' since measures must be countable additive). Let $B(X)$ be the collection of all bounded real-valued functions on $X$. What I'm trying to do here is to define the integral of $f \in B(X)$ with respect to $\mu$. Notice that every function $f \in B(x)$ is 'measurable'.
First define the integral for simple functions in $B(X)$. It is perfectly fine to do so. But how should we define the integral for for arbitrary $f \in B(X)$? I think the only reasonable way here is to define upper and lower integrals independently and show that they always coincide. Using upper and lower integrals, we can easily show that integral is additive. Can something similar be done for Lebesgue integrals?
To summarize, can the additivity of the lebesgue integral be established without any use of convergence theorems?; preferably using the finite additivity of the measure $\mu$?