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

Borel set

Def. Given a T.S. $(X, T)$, we call $\mathfrak{B}=\cap_{T \subseteq \mathfrak{M}}\mathfrak{M}$, the smallest $\sigma$-algebra containing $T$ Borel $\sigma$-algebra of $X$ and the element of it is called by a Borel set.

NB1; By definition, every open set and closed set are a Borel set.
NB2; $G_\delta$-set (countable intersection of open set) $\cap_{n=1}^\infty U_n$ is Borel set.
NB3; $F_\sigma$ set (countable union of closed set) $\cup_{n=1}^\infty F_n$ is Borel set.
NB4; Thus, differently from topology, Borel set is closed under countable intersection.
NB5; Given $(\mathbb{R}, T_{std})$, $[ a, b ) = \cap_{n=1}^{\infty} (a-\frac{1}{n}, b)$ is Borel set.

Def. Given T.S. $(X, T_X), (Y, T_Y)$ and $f: (X, \mathfrak{B}_X)\rightarrow (Y, T_Y)$ is called Borel measurable function if $\forall U \in T_Y, f^{-1}(U) \in \mathfrak{B}_X$.

NB6; If $f: X\rightarrow Y$ is cts, then $f$ is Borel measurable, since every inverse image of open set in continuous function is open, i.e., Borel set.

Propositions

\[\text{Given a map } f: (X, \mathfrak{M}) \rightarrow (Y, T_Y), \nonumber\]

1. Let $\emptyset \ne \Omega := \{ E (\subseteq Y) : f^{-1} \in \mathfrak{M}\}$. Then, $\Omega$ is $\sigma$-algebra of $Y$.
2. If $f$ is measurable and $E \in \mathfrak{B}_Y$, then $f^{-1}(E) \in \mathfrak{M}$.
3. Let $Y=(\bar{\mathbb{R}}, T_{std})$, then $f$ is measurable iff $\forall \alpha \in \mathbb{R}, f^{-1}((\alpha, \infty] ) \in \mathfrak{M}$.
4. Let $f$ is measurable and $g:(Y,T_Y) \rightarrow (Z, T_Z)$ is Borel measurable. Then, $h=g\circ f$ is measurable.

Proof

1. For the inclusion relation of inverse function, check previous propositions.
   • a. Trivially, $Y, \emptyset \in \Omega$
   • b. Let $\forall n \in \mathbb{N}, E_n\in \Omega$. Since $f^{-1}(\cup_{n=1}^\infty E_n) = \cup_{n=1}^\infty f^{-1}(E_n) \in \mathfrak{M}$, $\cup_{n=1}^\infty E_n \in \Omega$.
   • c. Let $E \in \Omega$. Then, $f^{-1}(E^c) = (f^{-1}(E))^c \in \mathfrak{M}$.
2. By (1), $\Omega$ is $\sigma$-algebra of $Y$ and let $\mathfrak{B}_Y$ be a Borel $\sigma$-algebra of $Y$. Then, $E \in \mathfrak{B}_Y \subset \Omega$. By definition of $\Omega$, $f^{-1}(E) \in \mathfrak{M}$.
3. • $(\Rightarrow)$ $\forall \alpha \in \mathbb{R}, (\alpha, \infty] = \cup_{n=1}^\infty (\alpha, \alpha+n) \in \mathfrak{B}_Y $. By (2), if $f: X\rightarrow Y$ is measurable, $f^{-1}((\alpha, \infty])$ is measurable set. $\Leftrightarrow f^{-1}([-\alpha, \alpha ]) \in \mathfrak{M}$. Similarly, $f^{-1}([-\infty, \alpha )) \in \mathfrak{M} $$\Leftrightarrow f^{-1}([\alpha, \infty ]) \in \mathfrak{M}$.
   • $(\Leftarrow)$ $\forall \alpha \in \mathbb{R}, (f^{-1}((\alpha, \infty ]))^c = f^{-1}((\alpha, \infty ]^c) = f^{-1}([ -\infty, \alpha ])$. $\implies \cup_{n=1}^\infty f^{-1}[-\infty, \alpha - \frac{1}{n}] = f^{-1}(\cup_{n=1}^\infty [-\infty, \alpha - \frac{1}{n}] ) = f^{-1}([-\infty, \alpha )) \in \mathfrak{M} $. For any open set $(a,b)$, $f^{-1}((a,b)) = f^{-1}((a, \infty]) \cap f^{-1}([-\infty, b)) \in \mathfrak{M}$. Thus, $f$ is measurable. Then, it remains to show that any open set $U$ is a countable union of disjoint open intervals, which will be proved below.
4. $h^{-1} = f^{-1}\circ g^{-1}$. By definition, $\forall U \in T_Z, g^{-1}(U) \in \mathfrak{B}_Y$. Then, (2) concludes the proof.

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\[\text{Every open subset of } \mathbb{R} \text{ is a countable union of disjoint open intervals.} \nonumber\]

Proof

Originally this proof comes from stackexchange-mathematics.
Let $U$ be a non-empty open subset of $\mathbb{R}$. For $x, y \in U$, define $x \sim y$ iff $[\min(x,y), \max(x,y) ] \subseteq U$. Then, $\sim$ is an equivalence relation on $U$, since followings hold:

1. $[ x, x ] \subseteq U$, i.e., $x\sim x$
2. $[ \min(x,y), \max(x,y) ]= [ \min(y,x), \max(y,x) ] \subseteq U$.
3. Let $[ \min(x,y), \max(x,y) ]\subseteq U$ and $[ \min(y,z), \max(y,z) ]\subseteq U$. Since $[ \min(x,z), \max(x,z) ] \subseteq [ \min(x,y), \max(x,y) ] \cup [ \min(y,z), \max(y,z) ]$ holds, $x\sim z$.

Note that equivalence classes $R$ are pairwise disjoint open intervals in $\mathbb{R}$. Then, $U=\cup_{I \in R}I$ holds. It remains to show that $R$ is countable. For each $I \in R$, we can choose a rational number $q_I$ since each $I$ is disjoint, which means that they must contain distinct rationals. Therefore, the map $R\rightarrow \mathbb{Q} (I \mapsto q_I)$ is injective, which shows the countability.

Limit and measurable

Let $\{ f_n \} _{n=1}^\infty$ be a seq. of functions, where $\forall n, f _n: E \rightarrow \mathbb{R}, \emptyset \ne E \subseteq \mathbb{R}$. Define $\sup _{n \in \mathbb{N}} f _n(x) := \sup \{ f _n(x) : n\in \mathbb{N}\}$ and $\inf _{n \in \mathbb{N}} f _n(x) := \inf \{ f _n(x) : n\in \mathbb{N}\}$. Then, $\sup _{n \in \mathbb{N}} f _n(x)$ and $\inf _{n \in \mathbb{N}} f _n(x)$ are also a function from $E$ to $\bar{\mathbb{R}}$.

Propositions

\[\text{If } \forall n\in \mathbb{N},\ f_n: (X, \mathfrak{M}) \rightarrow \bar{\mathbb{R}} \text{ is measurable, then } \sup f_n , \inf f_n \text{ are measurable.} \nonumber\]

Proof

$(\sup f_n)^{-1}(\alpha, \infty] = \cup_{n=1}^\infty f_n^{-1}(\alpha, \infty]$. By above proposition 3, $\forall n \in \mathbb{N}, f_n^{-1}(\alpha, \infty] \in \mathfrak{M}, \forall \alpha \in \mathbb{R}$. Since $\mathfrak{M}$ is $\sigma$-algebra, $(\sup f_n)^{-1}(\alpha, \infty] \in \mathfrak{M}$, thus $\sup f_n$ is measurable. Similar argument holds for $\inf f_n$ by changing union to intersection.

NB1; From this, we can see that for measurable functions $f_n$, $\inf\sup f_n$ and $\sup\inf f_n$ are measurable.
NB2; For measurable functions $f_n$, suppose $f(x) = \lim_{n\rightarrow \infty}f_n(x)$ is well-defined for all $x$. Then, $f$ is measurable since $\lim f_n$ is defined by $\lim\inf$ and $\lim\sup$ which are measurable.

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Def. Let $f: X \rightarrow \bar{\mathbb{R}}$ be a measurable function. Define the positive part of $f$, $f^{+} := \max \{ f(x), 0 \} : X \rightarrow [0, \infty]$ and the negative part of $f$, $f^{-} := - \min \{ f(x), 0 \} : X \rightarrow [0, \infty]$.

NB3; $f = f^{+} - f^{-}$.

Propositions

\[\text{For measurable function } f: (X, \mathfrak{M}) \rightarrow \bar{\mathbb{R}} , \text{ it holds that } \nonumber\]

1. $f^{+}$ and $f^{-1}$ are measurable.
2. $| f | = f^{+} + f^{-}$ is measurable.

Proof

1. It suffices to show that pointwise maximum of measurable function is measurable since $0: X \rightarrow 0$ is a measurable function. For all $a \in \mathbb{R}$, $\{x\in X: \max(f(x), g(x)) > a \} = \{x\in X: f(x) > a \} \cup \{x\in X: g(x) > a \}$. Since $\mathfrak{M}$ is closed under countable union, $\{x\in X: \max(f(x), g(x)) > a \} \in \mathfrak{M}$, which conludes the proof.
2. Similarly, for all $a \in \mathbb{R}$, $\{x\in X: |f(x)| > a \}=\{x\in X: f(x) > a \} \cup \{x\in X: f(x) < -a \}$ holds, thus $|f |$ is measurable.

Simple function

Def. Given $X$, let $Y = [0, \infty)$. We say $s: (X, \mathfrak{M}) \rightarrow Y$ is simple if $| \text{Im}s| < \infty$, i.e., $\text{Im}s = \{a_1, \ldots, a_n \} \subseteq Y$.

NB4; Let $\text{Im}s= \{ a_1, \ldots, a_n\}, a_i \ne a_j$ if $i\ne j$. Then, $s^{-1}(\{a_i\})\cap s^{-1}(\{a_j\}) = \emptyset$ if $i\ne j$. Denote $A_i := s^{-1}(\{a_i\})$, then $S=\sum_{k=1}^n a_k \chi_{A_k}$, i.e., every simple function is a finite linear combination of charateristic function.
NB5; Let $s$ be simple. Then, $s$ is measurable iff $\forall i, A_i \in \mathfrak{M}$. (This can be proved by rm27-1)

Simple approximation theorem

\[\text{Let } f: (X, \mathfrak{M})\rightarrow [0, \infty] \text{ is measurable. Then, } \exists \{ s_n \}_{n=1}^\infty \text{ on } X, \text{ s.t. } \nonumber\]

1. $\forall x \in X$, $0\leq s_1(x) \leq s_2(x) \leq \ldots$, i.e., $\{ s_n(x)\}$ is monotone increasing sequence.
2. $\forall x\in X$, $s_n(x) \stackrel{n \rightarrow \infty}{\rightarrow} f(x)$.

Proof

• First, aim to construct a Borel measurable function $\varphi_n : [0, \infty] \rightarrow \mathbb{R}$ for each $n\in \mathbb{N}$. Then, fix $t \in (0, \infty)$ and $n\in \mathbb{N}$. Then, $\exists! k(n,t) \in \mathbb{Z}$ s.t. $k \leq 2^n t < k+1$ by density in $\mathbb{R}$. Define $\varphi_n : t \mapsto \begin{cases} \frac{k(n,t)}{2^n} \text{ if } t < n \\ n \quad n \leq t \leq \infty \end{cases},$ which is a Borel measurable function since $\varphi_n = n \chi_{[n, \infty]}+ \frac{k(n,\cdot)}{n^n}\chi_{[0, n)}$. Observe that $t-\frac{1}{2^n} < \varphi_n(t) \leq t, \forall n \in \mathbb{N}, \forall t\in[0, n]$ since $t < \frac{k+1}{2^n}$. Hence, $\{ \varphi_n \}$ is mononotonic increasing and $\varphi_n(t) \rightarrow t$ as $n \rightarrow \infty, \forall t \in [0, \infty]$.
• Then, define $s_n := \varphi_n \circ f (x \mapsto \varphi_n(f(x)))$, which is measurable simple function by Prop 4. Then, $\{ s_n \}$ is also monotonic increasing and $s_n(x) \stackrel{n \rightarrow \infty}{\rightarrow} f(x)$.