Random math: 29

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


Def. Given a measurable space $(X, \mathfrak{M})$, if $\mu: \mathfrak{M} \rightarrow [0, \infty]$ is a function satisfying countable additivity, then we call $\mu$ a measure on $X$, and $(X, \mathfrak{M}, \mu)$ is called measure space (M.S.)

NB1; Countable additivity means for any disjoint countable collection $\{ A _i \} _{i=1}^\infty \subseteq \mathfrak{M},\ \mu(\cup _{i=1}^\infty A _i) = \sum _{i=1}^\infty \mu(A _i)$
NB2; If a measure $\mu$ satisifes $\mu(X)=1$, we usually call $(X, F, \mu)$ a probability space and $\mu$ a probability measure.


\[\text{Given any measure } \mu: \mathfrak{M} \rightarrow [0, \infty] \text{, the following holds: } \nonumber\]
  1. $\mu(\emptyset) = 0$.
  2. If $\{ A_i \}_{i=1}^n$ is mutually disjoint, then $\mu(\cup _{i=1}^n A _i) = \sum _{i=1}^n \mu(A _i)$, i.e., $\mu$ is finite additive.
  3. If $A \subset B$ for $A, B \in \mathfrak{M}$, then $\mu(A) \leq \mu(B)$, i.e., monotonicity.
  4. If $A_n \subseteq A_{n+1}$ for $\forall n\in\mathbb{N}, A_n \in \mathfrak{M}$, let $A=\cup_{i=1}^\infty A_n \in \mathfrak{M}$. Then, $\mu(A) = \lim_{n\rightarrow \infty}\mu(A_n)$, i.e., continuity.
  5. If $B_n \supseteq B_{n+1}$ for $\forall n\in\mathbb{N}, B_n \in \mathfrak{M}$, let $B=\cap_{i=1}^\infty B_n \in \mathfrak{M}$ and for some $k \in \mathbb{N}, \mu(B_k) < \infty$. Then, $\mu(B) = \lim_{n\rightarrow \infty}\mu(B_n)$.


  1. Choose $A \in \mathfrak{M}$ s.t. $\mu(A) < \infty$. Then, $\mu(A\cup \emptyset \cup \emptyset \ldots \cup \emptyset) = \mu(A) + \sum_{n=1}^\infty \mu (\emptyset)=\mu(A)$, which implies $\mu(\emptyset)=0$.
  2. Take $A_{n+1} = A_{n+2} = \ldots = \emptyset$. Then, $\mu(\cup_{i=1}^n A_i \cup \emptyset \ldots)= \sum _{i=1}^n \mu(A _i) +0$.
  3. $\mu(B) = \mu(A \cup (B-A)) \stackrel{2}{=} \mu(A) + \mu(B-A)$. Since $\mu(B-A) \geq 0$, $\mu(A) \leq \mu(B)$.
  4. Define $E _n = A _n \setminus A _{n+1} = A _n \cap A _{n+1}^c \in \mathfrak{M}$, then $\{ E _n \} _{n=1}^\infty \subseteq \mathfrak{M}$, $\cup _{n=1}^\infty E _n = A$ and $\cup _{k=1}^n E _k = A _n$. $\mu(A) = \mu(\cup _{k=1}^\infty E _k ) = \sum _{k=1}^\infty \mu (E _k)=\lim _{n\rightarrow \infty} \sum _{k=1}^n \mu(E _k) \stackrel{2}{=} \lim _{n\rightarrow \infty} \mu( \cup _{k=1}^n E _k ) =\lim _{n\rightarrow \infty} \mu( A _n )$.
  5. Let $B_k \in \mathfrak{M}$ s.t. $\mu(B_k) < \infty$. Define $F_n := B_k \setminus B_n$ for $n \geq k$. Then, $B_n \supseteq B_{n+1} \Leftrightarrow B_n^c \subseteq B_{n+1}^c \leadsto B_k \cap B_n^c\subseteq B_k \cap B_{n+1}^c$, i.e., $F_n \subset F_{n+1}$. By 4, $\lim_{n\rightarrow \infty} \mu (F_n) = \mu (\cup_{n=1}^\infty F_n) = \mu (\cup_{n=1}^\infty B_k \cap B_n^c) = \mu (B_k - \cap_{n=1}^\infty B_n) = \mu(B_k) - \mu (\cap_{n=1}^\infty B_n)$. Here, $\lim_{n\rightarrow \infty} \mu (F_n)= \lim_{n\rightarrow \infty} \mu (B_k) - \mu(B_n) = \mu(B_k) - \lim_{n\rightarrow \infty} \mu (B_n)$.


Def. Given a measure space $(X, \mathfrak{M}, \mu)$,

  1. Let $s: X \rightarrow [0, \infty]$ be measurable simple function, i.e., $s = \sum_{i=1}^n \alpha_i \chi_{A_i}, A_i \in \mathfrak{M}$. Define $\int_{E} s \mathrm{d}\mu := \sum_{i=1}^n \alpha_i \mu(A_i \cap E), E \in \mathfrak{M}$.
  2. Let $f: X \rightarrow [0, \infty]$ be measurable function. Define $\int_{E} f \mathrm{d}\mu := \sup \{ \int_{E} s \mathrm{d}\mu : s(x) \leq f(x), \forall x \in E\}$.
  3. Let $f: X \rightarrow [ -\infty, \infty]$ be measurable function. Decompose $f = f^{+} - f^{-}$ and define $\int_{E} f \mathrm{d}\mu:= \int_{E} f^{+} \mathrm{d}\mu - \int_{E} f^{-} \mathrm{d}\mu, \forall E \in \mathfrak{M}$.


  1. $0 \leq f \leq g : X \rightarrow [0, \infty]$, then $\int_{E} f \mathrm{d}\mu \leq \int_{E} g \mathrm{d}\mu, \forall E \in \mathfrak{M}$.
  2. $A \subseteq B, f \geq 0$, then $\int_{A} f \mathrm{d}\mu \leq \int_{B} f \mathrm{d}\mu$.
  3. $\forall c \in \mathbb{R}, c\int_{E} f \mathrm{d}\mu = \int_{E} cf \mathrm{d}\mu$.
    Let $f: X \rightarrow [-\infty, \infty]$ be measurable on $E \in \mathfrak{M}$.
  4. If $f=0$, then $\int_{E} f \mathrm{d}\mu = 0$ even if $\mu(E)= \infty$.
  5. If $\mu(E)= 0$, then $\int_{E} f \mathrm{d}\mu = 0$ even if $f=\infty$.
  6. $\forall E \in \mathfrak{M}, \int_{E} f \mathrm{d}\mu= \int_{X} \chi_E f \mathrm{d}\mu$.

\[\text{Given M.S. } (X, F, \mu), \text{ let } s,t : X \rightarrow [0, \infty] \text{ be measurable simple functions.} \nonumber\] \[\text{Define } \varphi_a :F \rightarrow [0, \infty] \left(E \mapsto \int_{E} a \mathrm{d}\mu \right), \text{for } a \in \{s, t\}.\text{ Then, } \varphi_a \text{ is a measure and } (\varphi_s + \varphi_t)(E) := \varphi_s(E) + \varphi_t(E). \nonumber\]


  1. Let $\{ E _i \} _{i=1}^\infty$ be mutually disjoint measurable sets. $\varphi _s ( \cup _{i=1}^\infty E _i ) = \int _{\cup _{i=1}^\infty E _i} s \mathrm{d}\mu= $$\sum _{finite \ j} \alpha _j \mu (A _j \cap (\cup _{i=1}^\infty E _i)) $ $=\sum _{finite \ j} \alpha _j \sum _{i=1}^\infty \mu (A _j \cap E _i)$ $= \sum _{i=1}^\infty \sum _j \alpha _j \mu (A _j \cap E _i) =\sum _{i=1}^\infty \varphi _s (E _i)$. Thus, $\varphi _s$ and $\varphi _t$ are measures.
  2. Let $s = \sum_j \alpha_j \chi_{A_j}$ and $t=\sum_l \beta_l \chi_{B_l}$ and $E_{jl} = A_j \cap B_l$. Then, $\int_{E_{jl}} s \mathrm{d}\mu + \int_{E_{jl}} t \mathrm{d}\mu = \sum_j \alpha_j \mu(E_{jl})+ \sum_l \beta_l \mu(E_{jl})$. And $\int_{E_{jl}} s+t \mathrm{d}\mu = \sum_{j,l} (\alpha_j + \beta_l) \mu (E_{jl})$, which concludes the proof since $X$ is disjoint union of $E_{jl}$.

Lebesgues’s monotone convergence theorem

\[\text{Let } \{ f_n \}_{n=1}^\infty \text{ be a monotone increasing seq. of non-negatvie measurable function on M.S. } (X, F, \mu) \ \forall x \in X. \nonumber\] \[\text{Then, } \exists! f: X \rightarrow [0, \infty] \text{ s.t. } \forall x \in X, \ f(x) = \lim_{n\rightarrow \infty} f_n(x) \text{ and } \lim_{n\rightarrow \infty}\int_{X} f_n \mathrm{d}\mu = \int_{X} \lim_{n\rightarrow \infty} f_n \mathrm{d}\mu = \int_{X} f \mathrm{d}\mu. \nonumber\]

As a corollary, let $f:X \rightarrow [0, \infty]$ be measurable. Then, by simple approximation thm, there exists a monotone increasing seq. of simple measurable functions $\{ s _n \} _{n=1}^\infty$ s.t. $0 \leq s _n \leq f$. By LMCT, $\int _{X} f \mathrm{d}\mu = \lim _{n\rightarrow \infty}\int _{X} s _n \mathrm{d}\mu $ holds.


The existence of pointwise limit is obvious by MCT, i.e., $f(x) = \lim _{n\rightarrow \infty} f _n(x)$. Thus, we need to show $ \lim _{n\rightarrow \infty}\int _{X} f _n \mathrm{d}\mu = \int _{X} f \mathrm{d}\mu$.

  1. $( \lim _{n\rightarrow \infty}\int _{X} f _n \mathrm{d}\mu \leq \int _{X} f \mathrm{d}\mu)$: By definition, $\forall n \in \mathbb{N}, \ x \in X, \ f _n(x) \leq f(x)$ holds, i.e, $\forall n \in \mathbb{N}, \ \int _{X} f _n \mathrm{d}\mu \leq \int _{X} f \mathrm{d}\mu)$. Since $\{ f _n \} _{n=1}^\infty$ is monotone increasing seq., $\lim _{n\rightarrow \infty}\int _{X} f _n \mathrm{d}\mu \leq \int _{X} f \mathrm{d}\mu$ holds.
  2. $( \lim _{n\rightarrow \infty}\int _{X} f _n \mathrm{d}\mu \geq \int _{X} f \mathrm{d}\mu)$: Let $s$ be simple measruable function such that $0 \leq s \leq f$ and take $0 < c < 1$. Define mesurable sets $ E _n := \{ x\in X : f _n (x) \geq c s(x) \}$. Then, $E _n \subseteq E _{n+1}$ holds since $f _{n+1}(x) \geq f _n(x) \geq cs(x)$ and $X=\cup _{n=1}^\infty E _n$ holds since for any $x\in X$, if $f(x)=0$, then $x \in E _1$ or if $f(x)>0$, then $cs(x) < f(x)$ since $c<1$, hence $\exists n \in \mathbb{N}$ s.t. $x \in E _n$. Thus, $a _n := \int _{X} f _n \mathrm{d}\mu \stackrel{X \supseteq E _n}{\geq} \int _{E _n} f _n \mathrm{d}\mu \geq \int _{E _n} cs \mathrm{d}\mu \stackrel{c>0}{=} c\int _{E _n} s \mathrm{d}\mu =: c \varphi _s (E _n)$. Here, $\{a _n\}$ is a monotone increasing seq., thus $\exists \lim a _n$ by MCT and the same holds for $\varphi _s (E _n)$. Therefore, $\lim a _n \geq c \lim \varphi _s (E _n)$ and by continiuity of measure, it holds that $\lim \varphi _s (E _n) = \varphi _s \cup _{n=1}^\infty (E _n) = \varphi _s (X)$. Thus, $\lim _{n\rightarrow \infty} \int _{X} f _n \mathrm{d}\mu \geq c\int _{X} s \mathrm{d}\mu$ holds for arbitrary $c \in (0,1)$, which leads to $ \int _{X} s \mathrm{d}\mu \leq \lim _{n\rightarrow \infty} \int _{X} f _n \mathrm{d}\mu$. By definition of $s$, $0 \leq s \leq f$, this gives that $\int _{X} f \mathrm{d}\mu \leq \lim _{n\rightarrow \infty} \int _{X} f _n \mathrm{d}\mu$.

\[\text{If } f_n : X\rightarrow [0, \infty] \text{ is measurable } \forall n \in \mathbb{N}, \text{ and } f(x) := \sum_{n=1}^\infty f_n(x). \text{ Then, } \int _{X} f \mathrm{d}\mu = \sum _{n=1}^\infty \int _{X} f _n \mathrm{d}\mu. \nonumber\]


By simple approximation thm, there exist monotone increasing seqs. of simple measurable functions $s_i^n$ for each $f_n$ such that $s_i^n \rightarrow f_n$. By above prop and LMCT, it holds that $\int _X f _1 + f _2 \mathrm{d}\mu = \int _{X} f _1 \mathrm{d}\mu + \int _{X} f _2 \mathrm{d}\mu$. Similarly, the same will hold for $g _N = f _1 + \cdots + f _N$ that $ \int _X g \mathrm{d}\mu = \sum _{n=1 }^N \int _{X} f _n \mathrm{d}\mu$. Applying MCT conlcudes the proof.

Fatou’s Lemma

\[\text{If }\forall n \in \mathbb{N}, f_n : X \rightarrow [0, \infty] \text{ is measurable, then } \int_X \liminf_{n\rightarrow \infty} f_n \mathrm{d}\mu \leq \liminf_{n\rightarrow \infty} \int_X f_n \mathrm{d}\mu. \nonumber\]


Define $g_n(x) := \inf_{k\geq n}f_k(x)$. Then, $g_n(x) \geq f_n(x)$ and $\{g_n\}$ is a monotone increasing sequence such that $g_n(x) \rightarrow \liminf f_n(x)$. Thus, $\int_X g_n \mathrm{d}\mu \leq \int_X f_n \mathrm{d}\mu$ holds, which gives that $\liminf \int_X g_n \mathrm{d}\mu \leq \liminf \int_X f_n \mathrm{d}\mu$. Here, $\liminf \int_X g_n \mathrm{d}\mu \stackrel{MCT}{=} \lim \int_X g_n \mathrm{d}\mu \stackrel{LMCT}{=} \int_X \lim g_n \mathrm{d}\mu = \int_X \liminf f_n \mathrm{d}\mu$ holds, which concludes the proof.