Measures

June 13, 2024

There is no way of measuring the importance of measure.

Definition 1.
Let $\mathcal{F}$ be a $\sigma$-algebra on $\Omega$. A function $\mu:\mathcal{F}\to[0,+\infty]$ is a measure if

$\mu(\emptyset)=0$.
$\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$ for all $A_n\in\mathcal{F}$ where $A_i\cap A_j=\emptyset$ for $i\neq j$.

If $\mu(\Omega)\lt +\infty$, then $\mu$ is a finite measure.
If $\mu(\Omega)=1$, then $\mu$ is a probability measure.

Definition 2.
Let $\mu$ be a non-negatice set function over $\mathcal{C}$. Then $\mu$ is

finite addable, if $A_i\in\mathcal{C}$ are disjoint and $\sum_{i=1}^n A_i\in\mathcal{C}$, then $\mu(\sum_{i=1}^n A_i)=\sum_{i=1}^n\mu(A_i)$.
$\sigma$-addable, if $A_i\in\mathcal{C}$ are disjoint and $\sum_{i=1}^{+\infty} A_i\in\mathcal{C}$, then $\mu(\sum_{i=1}^{+\infty} A_i)=\sum_{i=1}^{+\infty}\mu(A_i)$.
downward continuous, if for $A_i\in\mathcal{C}$ and $A_n\uparrow A\in\mathcal{C}$, $\lim_{n\to\infty}\mu(A_n)=\mu(A)$.
upward continuous, if for $A_i\in\mathcal{C}$ and $\mu(A_0)\lt +\infty$, $A_n\downarrow A\in\mathcal{C}$, $\lim_{n\to\infty}\mu(A_n)=\mu(A)$.
continuous upon zero, if for $A_i\in\mathcal{C}$ and $\mu(A_0)\lt +\infty$,$A_n\downarrow\emptyset$, $\lim_{n\to\infty}\mu(A_n)=0$.

Theorem 3. Let $\mu$ be a finite addable non-negative set function over an algebra $\mathcal{C}$. Then $\mu$ is $\sigma$-addable $\Leftrightarrow$
$\mu$ is downward continuous $\Rightarrow$ $\mu$ is upward continuous $\Rightarrow$ $\mu$ is continuous upon zero.

Proof.

$\mu$ is $\sigma$-addable $\Rightarrow$ $\mu$ is downward continuous:

Let $A_n\uparrow A$ and $B_n=A_{n+1}\setminus A_n$.
Then $A=\sum_{n=1}^{+\infty}A_n=\sum_{n=1}^{+\infty}B_n\bigcup A_1$.
$\mu(A)=\sum_{n=1}^{+\infty}\mu(B_n)+\mu(A_1)=\lim_{n\to\infty}\sum_{i=1}^{n}(\mu(A_{i+1})-\mu(A_i))+\mu(A_1)=\lim_{n\to\infty}\mu(A_n)$.

$\mu$ is downward continuous $\Rightarrow$ $\mu$ is $\sigma$-addable:

Let $A_i\in\mathcal{C}$ be disjoint and $\sum_{i=1}^{+\infty}A_i\in\mathcal{C}$.
Then $B_n=\sum_{i=1}^{n}A_i\uparrow B=\sum_{i=1}^{+\infty}A_i$.

$\mu$ is downward continuous $\Rightarrow$ $\mu$ is upward continuous:

Let $A_n\downarrow A$ with $\mu(A_1)\lt +\infty$.
Then $B_n=A_0\setminus A_n\uparrow A_0\setminus A$, where $A_0\in\mathcal{C}$, $A_0$ has finite measure and $A_1\subset A_0$.
Thus $\mu(A_n)=\mu(A_0)-\mu(B_n)\to\mu(A_0)-\mu(A_0\setminus A)=\mu(A)$.

$\mu$ is downward continuous $\Rightarrow$ $\mu$ is $\sigma$-addable:

Let $A_i\in\mathcal{C}$ be disjoint and $\sum_{i=1}^{+\infty}A_i\in\mathcal{C}$.
Then $B_n=\sum_{i=1}^{n}A_i\uparrow B=\sum_{i=1}^{+\infty}A_i$.

$\mu(B)=\lim_{n\to\infty}\mu(B_n)=\lim_{n\to\infty}\sum_{i=1}^{n}\mu(A_i)=\sum_{i=1}^{+\infty}\mu(A_i)$.

Theorem 4. If $\mu(\Omega)\lt +\infty$, then the above statements are equivalent.

Example 5. Let $\Phi(x)$ be the Cantor function over $[0,1]$, and
$\Psi(x)=\frac{x+\Phi(x)}{2}$. Let $C$ denote the Cantor set over $[0,1]$, and $W\subset\Psi(C)$ be a non-measurable set.
Let $g(x)=\Psi^{-1}(x), f(x)=\chi_{\Psi^{-1}(W)}$,
then $f\circ g$ is non-measurable.

If $x=\sum_{k=1}^{+\infty}\frac{a_k}{3^k}\in C$, then every $a_k$ is either $0$ or $2$ (we can choose a proper representation to achieve that, i.e. $0.1=0.022222\dots$). Then the cantor function is defined by
$\Phi(x)=\sum_{k=1}^{+\infty}\frac{a_k/2}{2^k}$.

Why add $x$? If so, $\Psi^{-1}(W)\subsetneq C$ thus measurable. If not, $\Psi^{-1}(W)$ can probably contain an
interval, and it’s not necessary to be measurable.

Measure Theory