====== Lecture 10 ======
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).