# Random math: 27

Source of most content: Gunhee Cho.

# Measurable set

Def. Given a set $X$, we call $\mathfrak{M} \subseteq 2^X$ a $\sigma$-algebra in $X$ if it satisfies

1. $X \in \mathfrak{M}$.
2. If $A \in \mathfrak{M}$, then $A^c \in \mathfrak{M}$.
3. If $\{ A _n \} _{n=1}^\infty \subseteq \mathfrak{M}$, then $\cup _{n=1}^\infty A _n \in \mathfrak{M}$, i.e., countable cases

Def. Given a set $X$, we call an element of $\sigma$-algebra a measurable set and we call $X$ a measurable space

NB1; In probability, we usually ues the term $\sigma$-field and event.
NB2; If we consider the finite union instead of countable union in 3, we call $\mathfrak{M}$ an algebra of $X$.
NB3; By 1 and 2, $X, \emptyset \in \mathfrak{M}$ and for $A, B \in \mathfrak{M}, A \cap B \in \mathfrak{M}$ and $A -B \in\mathfrak{M}$ by 2 and 3.
NB4; If $\{ A _n \} _{n=1}^\infty \subseteq \mathfrak{M}$, then $\cap _{n=1}^\infty \in \mathfrak{M}$.
NB5; In general, $\exists T_X$ on some set $X$ s.t. $| T_X | = | \mathbb{N} |$. However, $\not\exists \sigma$-algebra $\mathfrak{M}$ of $X$ s.t. $\mathfrak{M}$ is countable.

### Proof (NB5)

This implies that $\sigma$-algebra has either finite or uncountably infinite members and prove this by contradiction.
Assume there exists an infinitely countable $\sigma$-algebra $\mathfrak{M}$ on infinite set $X$. Define $f: X \rightarrow \mathfrak{M} (x \mapsto \bigcap_{A\in \mathfrak{M}, x \in A}A)$, i.e., the smallest measurable set which contains $x$. Then, $\text{Im} f$ will construct the partition of $X$ and every member of $\mathfrak{M}$ is a unique union of partitions. By assumption, $S$ has infinitely countable members, the partition has to be at least countably infinite. However, in this case, there are uncountably many possible unions of such sets, which is a contradiction. Thus, $\mathfrak{M}$ must be uncountable.

Def. Given a $\sigma$-algebra $\mathfrak{M}$ of $X$ and a T.S. $(Y, T_Y)$, a function $f: X \rightarrow Y$ is called a measurable function if $\forall V \in T_Y, f^{-1}(V) \in \mathfrak{M}$.

NB6; In measurable function, the inverse image of open set is a measuarble set, which is very similar to countinuity in topology.
NB7; Let $f: X \rightarrow Y$ is measurable and $g:Y \rightarrow Z$ is continous. Then, $g \circ f$ is measurable. $\because$ for open set $U$ in $Z$ $(g\circ f)^{-1}(U) = f^{-1}(g^{-1}(U)) \in \mathfrak{M}$.

## Propositions

$\text{Given measurable space } (X, \mathfrak{M}) \text{, T.S. } (Y, T_Y) \text{ and two measurable functions } u: X \rightarrow \mathbb{R}, v: Y \rightarrow \mathbb{R}, \nonumber$ $\text{Let } \Phi: \mathbb{R} \times \mathbb{R} \rightarrow Y \text{ be cts. Then, } h: X \rightarrow Y \ (x \mapsto \Phi(u(x)\times v(x))) \text{ is measurable.} \nonumber$

### Proof

From above NB7, it suffices to show $f: X \rightarrow \mathbb{R} \times \mathbb{R} (x \mapsto (u(x) \times v(x)))$ is measurable since $\Phi$ is cts. Take $(a_1, b_1) \times (a_2, b_2) \subseteq \mathbb{R} \times \mathbb{R}$. Then, $f^{-1}((a_1, b_1) \times (a_2, b_2)) = u^{-1}((a_1, b_1)) \cap v^{-1}((a_2, b_2)) \stackrel{mble \ u,v}{\in} \mathfrak{M}$.

$\text{By above proposition, for measurable space } (X, \mathfrak{M}), \nonumber$
1. $\Phi : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{C} ( (x, y) \mapsto x+iy)$ is continous. Thus, if $u, v : X \rightarrow \mathbb{R}$ are measurable, then $u + iv : X \rightarrow \mathbb{C}$ is measurable function.
2. Conversely, if $u + iv : X \rightarrow \mathbb{C}$ is measurable, since $Re: \mathbb{C} \rightarrow \mathbb{R} (x+iy \rightarrow x)$ and $Im: \mathbb{C} \rightarrow \mathbb{R} (x+iy \rightarrow y)$ are cts, $Re(u+iv) = u: X \rightarrow \mathbb{R}$ and $Im(u+iv) = v : X \rightarrow \mathbb{R}$ are measurable.
3. On $(\mathbb{R}, T_{std})$, $+: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ and $\times: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ are continous, thus if $u,v: X \rightarrow \mathbb{R}$ are measurable, then $u+v$, $uv$ are measurable.

$\text{Given } E \in \mathfrak{M}, \text{ define an indicator function } \chi_E : X\rightarrow \mathbb{R} (x \rightarrow \mathbf{1}_{x \in E}), \text{ Then, } \chi_E \text{ is measurable.} \nonumber$

### Proof

For any open $A$ in $\mathbb{R}$, $\chi_E^{-1}(A) \in \{ X, E, E^c, \emptyset \}$, where all elements are measurable. Thus, indicator function is measurable.

$\text{Let } f: X \rightarrow \mathbb{C} \text{ be measurable function. Then, } \exists \text{measurable } \alpha: X \rightarrow \mathbb{C} \text{ s.t. } |\alpha| =1 \text{ and } f = | \alpha | f. \nonumber$

### Proof

Let $E= f^{-1}(\{0\}) \in \mathfrak{M}$, where $\{ 0 \}$ is closed set in $\mathbb{R}^2$ and $\alpha: X \rightarrow \mathbb{C} ( x \mapsto \frac{f(x) + \chi_E(x)}{ | f(x) + \chi_E(x) | } )$. Then, if $x \in E$, i.e., $f(x)=0$, $\alpha(x) =1$ and if $x \in E^c, | \alpha(x) | = 1$ thus $\alpha(x) = \frac{f(x)}{|f(x)|}$. It remains to show $\alpha$ is measurable, which suffices to observe $\varphi: \mathbb{C}- \{ 0 \} \rightarrow \mathbb{C} (z \mapsto \frac{z}{|z|})$ is cts. Then, $\alpha = \varphi \circ (f + \chi_E)$, where $f$ and $\chi_E$ are measurable and thus its sum is also measurble. Finally, by NB7, $\alpha$ is measurable.

$\text{Given a set } X, \text{ let } F \subseteq 2^X. \text{ Then, } \exists \text{smallest } \sigma \text{-algebra } \mathfrak{M}^* \text{ of } X \text{ which contains } F. \nonumber$

### Proof

Let $\mathfrak{M}$ be any $\sigma$-algebra of $X$. Take $\mathfrak{M}^* := \cap_{F \subset \mathfrak{M}}\mathfrak{M}$. Then, $\mathfrak{M}^*$ is a $\sigma$-algebra.

1. $\forall \mathfrak{M}$, $X \in \mathfrak{M}$. Thus, $X \in \mathfrak{M}^*$.
2. Let $A \in \mathfrak{M}^*$. Then, $\forall \mathfrak{M}, A \in \mathfrak{M}$. Since $\mathfrak{M}$s are $\sigma$-algebra, $\forall \mathfrak{M}, A^c \in \mathfrak{M}$. Thus, $A^c \in \mathfrak{M}^ *$.
3. Let $\{ A_n \} _{n=1}^\infty$ be a seq. in $\mathfrak{M}^*$. Then, it is a seq. in all $\mathfrak{M}$. Since $\mathfrak{M}$s are $\sigma$-algebra, $\cup _{n=1}^\infty A _n \in \mathfrak{M}$. Thus, $\cup _{n=1}^\infty A _n \in \mathfrak{M}^ *$.

Let $\Sigma$ be a $\sigma$-algebra s.t. contians $F$ and assume $\Sigma \subset \mathfrak{M}^ *$. By definition of $\mathfrak{M}^ *$, this is a contradiction. Thus, $\mathfrak{M}^ *$ is a smallest $\sigma$-algebra which contains $F$.