We did Tao 8.2.
Main result is monotone convergence theorem: given a monotone increasing sequence of non-negative measurable functions $f_n$, we have $$ \int \lim f_n = \lim \int f_n$$ or equivalently
$$ \int \sup f_n = \sup \int f_n$$
The $\geq $ direction is easy, the $\leq$ direction is hard, which requires 3 steps lowering of the LHS $\int \sup f_n$:
We first replace $\sup f_n$ by simple functions $s$, with $\sup f_n \geq s$, for some simple function $s$ sub-ordinate to $\sup f_n$.
We then lower $s$ a bit, $ s \geq (1-\epsilon)s $.
We then cut-off the integration domain a bit, by introducing a cut-off function $1_{E_n}(x)$, where $E_n= \{x: (1-\epsilon)s(x) \leq f_n(x) \}$, we get $ (1-\epsilon)s \geq (1-\epsilon) s 1_{E_n}$.
After the three lowering, we get $(1-\epsilon) s 1_{E_n} \leq f_n$, hence
$$ \int (1-\epsilon) s 1_{E_n} \leq \int f_n \leq \sup \int f_n$$
Then, we reverse the above lowering process, by taking limit, or sup over all possible choices
First, we let $n \to \infty$. By proving directly a 'baby version' of monotone convergence theorem for simple functions, we have that $ \sup \int s 1_{E_n} = \int s \sup 1_{E_n} = \int s$. This gives us
$$\int (1-\epsilon) s \leq \sup \int f_n $$
Then, we take limit $\epsilon \to 0$, to get $$\int s \leq \sup \int f_n $$
Finally, we sup over all simple functions $s$ subordinate to $\sup f_n$, to get $$ \int \sup f_n \leq \sup \int f_n$$
Then, we did some applications. For example, summation and integration can commute now (for non-negative measurable functions).