# Random math: 31

Source of most content: Gunhee Cho and RCA by Rudin.

# Almost everywhere

Def. Given a M.S. $(X, F, \mu)$, we say a statement $P$ holds almost everywhere (a.e.) on $X$ if $\exists N \in F$ s.t. $\mu(N)=0$ and $P$ is true for $X\setminus N$.

NB1; Give the relation to measurable fns $f,g: X \to [0, \infty]$, $f \sim g$ if $\mu( \{x \in X : f(x)\ne g(x) \})=0$ holds, i.e., $f \sim g$ if $f=g$ a.e.. Then, $\sim$ is an equivalence relation.

1. $f\sim f$ since $\{x \in X : f(x)\ne g(x) \} = \emptyset$ and $\mu(\emptyset) = 0$.
2. If $f\sim g$, then $g \sim f$.
3. Suppose $f \sim g$ and $g\sim h$. Let $A= \{x \in X : f(x)\ne g(x) \}$ and $B= \{x \in X : g(x)\ne h(x) \}$. Since $\{x \in X : f(x)\ne h(x) \} \subset A\cup B$, $\mu(\{x \in X : f(x)\ne h(x) \}) \leq \mu(A\cup B) = \mu(A)+\mu(B) = 0$. Thus, $f\sim h$.

Let $E= \{ x \in X : f = g\}$. If $f \sim g$, then $\int_X f = \int_{E \cup E^c} f = \int_E f + \int_{E^c}f = \int_E g + \int_{E^c} f \stackrel{\mu(E^c)=0}{=} \int_E g$ $= \int_E g + \int_{E^c} g = \int_X g$. Hence, if $f\sim g$ then $\int_X f = \int_X g$. So, roughly speaking, we can neglect the measure zero set in integration.

# Completion

$\text{Given M.S. } (X,F,\mu), \text{ define } F^* := \{ E \subseteq X : \exists A,B \in F \text{ s.t. } A \subset E \subset B \text{ and } \mu(B-A) = 0 \} \text{ and } \nonumber$ $\mu^*:= F^ * \to [0, \infty] (E \mapsto \mu(A)) \text{, where } A \subset E \subset B. \text{ Then, } (X, F^ *, \mu^ *) \text{ is a M.S. called the completion of } (X,F,\mu). \nonumber$

NB2; This extended measure $\mu^*$ is called complete, since all subsets of measure zero sets are now measurable and $F^ *$ is called the $\mu$-completion of $F$.
NB3; Every measure can be completed, so, whenever it is convenient, we may assume that any given measure is complete.
NB4; For M.S. $(X, F, \mu)$, let $E \subseteq F, \, f: E \to (Y, T)$. We say $f$ is measruable function if $\forall V \in T, f^{-1}(V) \cap E \in F$ and $\mu (E^c) = 0$.

### Proof

• Well-definedness of $\mu^ *$: Let $E \in F^ *$ and $A_1 \subseteq E \subseteq B_1$ and $A_2 \subseteq E \subseteq B_2$, i.e., $\mu(B_1 - A_1) = \mu(B_2 - A_2) = 0$. Then, $A_1 - A_2 \subset E - A_2 \subset B_2 - A_2$ holds, which gives $\mu(A_1 - A_2)=0$. Similarly, $\mu(A_2 - A_1)=0$ holds. By using countable additivity of measure, we have $\mu (A_1) = \mu(A_1 \cap A_2) + \mu(A_1 \cap A_2^c)$ and $\mu (A_2) = \mu(A_1 \cap A_2) + \mu(A_2 \cap A_1^c)$, which gives $\mu (A_1) = \mu(A_2)$.
• $F^*$ is $\sigma$-algebra:
1. Since $F \subseteq F^ *, \, X \in F^ *$.
2. If $A \subseteq E \subseteq B$ then $B^c \subseteq E^c \subseteq A^c$. Since $\mu(A^c - B^c) = \mu(B-A)$, if $E \in F^ *$ then $E^c \in F^ *$ holds.
3. Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$. $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i \implies \cup _{i=1}^\infty A _i \subseteq \cup _{i=1}^\infty E _i \subseteq \cup _{i=1}^\infty B _i \in F$. Hence, $\mu( \cup _{i=1}^\infty B _i - \cup _{i=1}^\infty A _i) = \mu( \cup _{i=1}^\infty (B _i - \cup _{j=1}^\infty A _j)) \leq \mu( \cup _{i=1}^\infty (B _i-A _i)) \stackrel{( * )}{\leq}\sum _{i=1}^\infty \mu(B _i - A _i) =0$, i.e., $\cup _{i=1}^\infty E _i \in F ^*$.
• Countable additivity: Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$ be mutually disjoint subsets. By deifnition, $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i, \, \mu(B _i - A _i) =0$. Then, $\mu^ * (\cup _{i=1}^\infty E _i) = \mu (\cup _{i=1}^\infty A _i) = \sum _i \mu (A _i) = \sum _i \mu^ * (E _i)$.
• $(*)$: Let $F_i = E_i - \cup_{j=1}^{i-1} E_i$. Then, $F_i$’s are mutually disjoint and $\cup_{i=1}^\infty F_i = \cup_{i=1}^\infty E_i$. Hence, $\mu(\cup_{i=1}^\infty E_i) = \mu(\cup_{i=1}^\infty F_i) = \sum_i \mu(F_i) \stackrel{monotonicity}{\leq} \sum_i \mu(E_i)$.

## Proposition

$\text{Let } \{ f_n \}_{n=1}^\infty \text{ be measurable fns defined a.e. on } X \text{ to } \mathbb{C} \text{ s.t. } \sum_{n=1}^\infty \int_X |f_n| d\mu < \infty. \nonumber$ $\text{Then, } f(x) := \sum_{n=1}^\infty f_n(x) \text{ is defined a.e. and } \int_X f d\mu = \int_X \sum_{n=1}^\infty f_n d\mu = \sum_{n=1}^\infty \int_X f_n d\mu. \nonumber$

### Proof

Let $s_n$ be a set which defines $f_n : s_n \to \mathbb{C}$, i.e., $\forall n \, \mu(s_n^c) =0$. Let $S := \cap_{n=1}^\infty s_n$. Then, $\mu (s^c) = \mu (\cup_{n=1}^\infty s_n^c) \leq \sum_{n} \mu(s_n ^c)=0$. Define $\varphi: S \to [0, \infty] \, (x \mapsto \sum_{i=1}^\infty \mid f_n (x) \mid)$. Then, $\int _X \varphi d\mu = \int _S \varphi d\mu \stackrel{(p)}{=}$ $\sum _{i=1}^\infty \int _S \mid f _n \mid d\mu = \sum _{i=1}^\infty \int _X \mid f _n \mid d\mu < \infty$, i.e., $\varphi \in L^1(\mu)$. Let $E = \{ x \in S : \varphi(x) < \infty \}$. Then, $\mu (E^c) = 0$. If $x\in E$, it holds that $\mid f(x) \mid = \mid \sum _{n=1}^\infty f _n(x) \mid \leq \sum _{n=1}^\infty \mid f _n (x) \mid = \varphi(x) < \infty$. This implies that $f$ is defined on $E$, i.e., $f$ is defined a.e. on $X$ so that $f \in L^1(\mu)$. Next, define $g _n(x):= \sum _{k=1}^n f _k(x)$ on $E$. Then, $g _n \to f$ on $E$ (a.e. on $X$). By LDCT, we have $\lim _n \int _E g _n d\mu = \int _E \lim _n g _n d\mu = \int _E f d\mu$. This concludes the proof since $\mu(E^c) = 0$. (p)