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--- math105-s22:notes:lecture_10 [2022/02/18 14:36]
pzhou
+++ math105-s22:notes:lecture_10 [2026/02/21 14:41] (current)
@@ Line 1: / Line 1: @@
 ====== 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 $$ \int \lim f_n = \lim \inf f_n$, or equivalently $\int \sup f_n = \sup \int f_n$.
+We did Tao 8.2.
+Main result is monotone convergence theorem: given a monotone increasing sequence of non-negative measurable functions $f_n$, we have $$ \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).

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