Prove Lebesgue Integral is between measure of the space times the inf and sup of function I am trying to prove the following, and it seems so easy that I'm afraid I might be oversimplifying the proof:
Show that is $(X, S, \mu)$ is a measure space and $f:X \to[0,\infty]$ is $S$-measurable, then:
$u(X)\inf\limits_{X} f \leq \int f d\mu \leq u(X)\sup\limits_{X} f$
where $\int f d\mu$ is the Lebesgue integral.
The proof seems incredibly straightforward to me:
If $P = \{A_1,A_2,...A_n\}$ is a partition of X then:
(1)
$\int f d\mu = \sup \{\sum^{n}_{i=1}u(A_i)\inf\limits_{A_i}\ f\} \geq \sum^{n}_{i=1}u(A_i)\inf\limits_{X} f = \inf\limits_{X} f \sum^{n}_{i=1}u(A_i)=\inf\limits_{X} f \,u(X)$
and
(2)
$\int f d\mu = \sup \{\sum^{n}_{i=1}u(A_i)\inf\limits_{A_i}\ f\} \leq \sum^{n}_{i=1}u(A_i)\sup\limits_{X} f = \sup\limits_{X} f \sum^{n}_{i=1}u(A_i)=\sup\limits_{X} f \,u(X)$
I saw another proof elsewhere online that invoked the simple function construction of the Lebesgue integral and it struck me as overcomplicating the proof. Might I be missing something?
 A: For a nonnegative measurable function, the integral $\int_X f\, d\mu$ is defined as the supremum of integrals $\int_X \varphi \, d\mu$ where $\varphi$ ranges over all simple functions with $0 \leqslant \varphi \leqslant f$.  Since $\underset{X}\inf f \cdot \chi_X$ is a simple function satisfying this condition, the first inequality is immediate, viz.
$$\underset{X}\inf f \cdot \mu(X) = \int_X \underset{X}\inf f \cdot \chi_X \, d\mu \leqslant \int_X f \, d\mu,$$
where the left-hand equality comes directly from the definition of the integral of a simple function.
How is this "overkill" with respect to a proof that unnecessarily introduces a partition and asserts (erroneously) that
$$\int_X f \, d\mu \,\,\underset{?}=\,\, \sup \{\sum^{n}_{i=1}u(A_i)\inf\limits_{A_i}\ f\}  $$
Over what set is the supremum taken here? Even if correctly posed you would need to justify
$$\int_X f \, d\mu = \sup_{\{A_i\}} \int_X \sum_{i=1}^n \inf_{A_i}f \cdot \chi_{A_i} \, d\mu,$$
which is many steps removed from the simple function definition of the integral.
Under the basic definition using simple functions it is easy to prove that $f \leqslant g$ implies that $\int_Xf \leqslant \int_Xg$ and this leads to
$$\int_X f \, d\mu \leqslant \sup_X f \cdot \mu(X)$$
