We will cover Tao's 7.5 and 8.1 today. Here we will use Tao's definition of measurable set, and Lebesgue integration, which a priori is not the same as Pugh's.

Let $\Omega \In \R^m$ be measurable, and $f: \Omega \to \R^m$ be a function. If for all open sets $V \In \R^m$, we have $f^{-1}(V)$ being measurable, then $f$ is called a measurable function.

If $f: \Omega \to \R^m$ is continuous, then $f$ is measurable. Indeed, if $f^{-1}(V)$ is open in $\Omega$, then $f^{-1}(V)$ is an intersection of open subset $U \In \R^m$ and $\Omega$ (recall the definition of topology on $\Omega$), an intersection of two measurable sets.

Instead of checking on all open sets $V \In \R^m$, we can just check for all open boxes in $\R^m$. Since any open can be written as a countable union of open boxes.

A measurable function $f$, post compose with a continuous function $g$ is still measurable. Since $$(g \circ f)^{-1}(open) = f^{-1} (g^{-1}(open)) = f^{-1}(open) = measurable $$

Lemma: $f: \Omega \to \R$ is measurable if and only if for all $a \in \R$, $f^{-1}( (a, \infty))$ is measurable.
Proof: every open set in $\R$ is a countable union of open interval, hence suffice to show that all open intervals $(a,b)$ has pre-image being measurable. We can easily show that $f^{-1}((a, b])$ is measurable for all $a<b$, and we can use countable operations to approximate open interval by half-open-half-closed ones, $(a,b) = \cup_{n} (a, b-1/n]$.

Simple functions are measurable functions $f: \Omega \to \R$, which takes value in a finite subset of $\R$.

Simple functions forms a vector space (i.e., closed under addition and scalar multiplication), and can be written as a finite linear combination of characteristic functions $\chi_E$.

The important thing is that, any non-negative measurable function $f$ admits a sequence of simple functions $f_n$, non-negative, and $f_n \leq f_{n+1}$, such that $f_n \to f$ pointwise. The construction requires both 'trunction' and refinement.

We then define integration for simple functions. Integration is a linear map from the vector space of simple function to $\R$.

Finally, in 8.2, we will define integration for non-negative measurable functions.

$\int f = \sup \{ \int s \mid 0 \leq s \leq f, \text{$s$ is a simple function \} $

For $f,g : \Omega \to [0, \infty]$, how to prove $\int f+g = \int f + \int g$?