This is how the integral on a measure space is defined in Ziemer's Modern Real Analysis textbook:
Let $(X, \mathcal{A}, \mu)$ be a measure space. We say that a function $f: X \to [-\infty, \infty]$ is countably simple if it is measurable and $f(X)$ is at most countable.
Given a non-negative countably simple function $f: X \to [-\infty, \infty]$ we define the integral of $f$ as $\int_X f d\mu = \sum_{i=1}^\infty a_i \mu( f^{-1}(\{ a_i \}))$ where $f(X) = \{ a_1, a_2, …\}$.
Given a countably simple function $f: X \to [- \infty, \infty]$ (not necessarily non negative), if at least one of the integrals $\int_X f^+ d\mu$, $\int_X f^- d\mu$ is finite, we say that the integral of $f$ exists and we write $\int_X f d\mu = \int_X f^+ d\mu – \int_X f^- d\mu$. Now if $f: X \to [-\infty, \infty]$ is any function, we define its upper integral
$$\overline{\int_X} f d\mu = \inf \{ \int_X g d\mu : g \mbox{ measurable, countably simple, and } g \geq f
\mbox{ a.e. }\}$$
and its lower integral
$$\underline{\int_X} f d\mu = \sup \{ \int_X g d\mu : g \mbox{ measurable, countably simple, and } g \leq f
\mbox{ a.e. }\} \; .$$
We say that the integral of $f$ exists if there values are equal and we say that $f$ is integrable if these numbers are equal to a finite value.
How is this definition equivalent to the standard one (by the standard definition i mean this: integration of simple measurable functions –> integration of non-negative functions by taking the supremum -> integration of arbitrary measurable functions using $f = f^+ – f^-$)?
Best Answer
Let's call this integral $I$, and the usual Lebesgue integral $\int$. I will suggest how to show $I(f) = \int f$ for measurable $f \colon X \to [0, \infty]$.
If you can prove that $I$ satisfies the monotone convergence theorem, then you are left with proving that $I(f) = \int f$ for measurable $f : X \to [0, \infty)$ with finite range (i.e. simple) since any measurable $f \colon X \to [0, \infty]$ is a pointwise increasing limit of measurable functions $f_n \colon X \to [0, \infty)$ with finite range. If you can prove that $I(f + g) = I(f) + I(g)$ and $I(cf) = cI(f)$ for $c \geq 0$, $f, g \colon X \to [0, \infty)$ measurable with finite range, then you are left with proving that $I(\chi_S) = \int \chi_S = \mu(S)$ for $S \in \mathcal{A}$.