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).