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?

• Where does $\int f d\mu = \sum^{n}_{i=1}u(A_i)\inf\limits_{A_i} f$ come from? Commented Oct 22, 2020 at 16:13
• @UmbertoP. Sorry you're right, there should be a $sup$ wrapped around that sum, will go edit now. I don't think that should change anything though? Commented Oct 22, 2020 at 16:17
• What you are doing now is using the simple function construction of the Lebesgue integral. Commented Oct 22, 2020 at 18:08
• @ChrisWang Sorry, what I mean is that the proof decomposed the simple function construction down to the level of the characteristic functions, which seems like overkill Commented Oct 22, 2020 at 18:31

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

$$\int_X f \, d\mu \leqslant \sup_X f \cdot \mu(X)$$