Measure on the Ring

Definition 1:

Let $X$ is a fixed set(or a Basic Space), and denote $\mathcal{E} := \left\{ E:E\subset X \right\}$. Call $\mathcal{E}$ is a set class on $X$.

Definition 2:

Let $\mathcal{E}$ is a set class on $X$, and if its elements satisfy:

$\forall E_1,E_2 \in \mathcal{E}$, have $E_1 \cup E_2 \in \mathcal{E}$ and $E_1-E_2 \in \mathcal{E}$, then we called $\mathcal{E}$ is a Ring on $X$ and denote: $\mathcal{R}$.

In particular, if we have $X \in \mathcal{R}$, we called $\mathcal{R}$ is a $\sigma-$Ring on $X$.

Definition 3:

Let $\mathcal{E}$ is a set class on $X$, and define the mapping from $\mathcal{E}$ to $X$:

$\mu : \mathcal{E}\to \mathbb{R}$

then, called $\mu$ is set function on $\mathcal{E}$.

Definition 4:

Let $\mathcal{R}$ is a Ring on $X$, and we define the set function $\mu : \mathcal{R}\to \mathbb{R}$, and satisfied:

i>. $\mu (\varnothing) = 0$

ii>. (Non-negative) $\forall E \in \mathcal{R}$, has $\mu (E) \ge 0$

iii>.(Additivity) $E_n \in \mathcal{R}$, $n=1,2,...$, and $\forall E_i \cap E_j=\varnothing (i \ne j)$, have:

$\mu \left(\bigcup_{n=1}^{\infty}E_n \right) =\sum_{n=1}^{\infty}\mu(E_n)$

then, we called $\mu$ is a measure on $\mathcal{R}$.

Theorem

If $\mathcal{R}$ is a Ring on $X$, and $\mu$ is a measure on $\mathcal{R}$, then it has:

  i>. (Limited additivity) If $E_1,E_2,...,E_n \in \mathcal{R}$, and $\forall E_i \cap E_j=\varnothing (i \ne j)$,have:

$\mu \left(\bigcup_{i=1}^{n}E_n \right) =\sum_{i=1}^{n}\mu(E_n)$

  ii>. (Monotonicity) If $E_1,E_2 \in \mathcal{R}$, and $E_1\subset E_2$, then

$\mu(E_1) \le \mu(E_2)$

  iii>. (Reducibility) If $E_1,E_2 \in \mathcal{R}$, $E_1\subset E_2$, and $\mu(E_1) \le \infty$, then

$\mu(E_2-E_1)=\mu(E_2)-\mu(E_1)$

  iv>. (Secondary additivity) If $E_n,E \in \mathcal{R}$, $n=1,2,...$, and $E\subset \bigcup_{n=1}^{\infty}E_n$, then

$\mu(E) \le \sum_{n=1}^{\infty}\mu(E_n)$

  v>. If $E_n \in \mathcal{R}$, , $E_1\subset E_2\subset \cdot \cdot \cdot ,$ and $\bigcup_{n=1}^{\infty}E_n \in \mathcal{R}$, then

$\mu \left(\bigcup_{n=1}^{\infty}E_n \right) =\lim_{n\to \infty}\mu(E_n)$

  vi>. If $E_n \in \mathcal{R}$, , $E_1\supset E_2\supset \cdot \cdot \cdot ,$ $\bigcap_{n=1}^{\infty}E_n \in \mathcal{R}$, and $\exists E_n$ such that $\mu(E_m) < \infty$, then

$\mu \left(\bigcap_{n=1}^{\infty}E_n \right) =\lim_{n\to \infty}\mu(E_n)$

In addition, if $\mathcal{R}$ is a $\sigma \text{-Ring}$, we have:

  vii>. If $E_n \in \mathcal{R}$, , then

$\mu\left(\lim\inf_{n}E_n\right) \le \lim\inf_{n}\mu(E_n)$

  viii>. If $E_n \in \mathcal{R}$, , and $\exists k \in \mathbb{N}$, such that $\bigcup_{n=k}^{\infty}E_n < \infty$, then

$\mu\left(\lim\sup_{n}E_n\right) \ge \lim\sup_{n}\mu(E_n)$

  ix>. If $E_n \in \mathcal{R}$, , $\lim E_n$ exists and $\exists k \in \mathbb{N}$ such that $\bigcup_{n=k}^{\infty}E_n < \infty$, then

$\mu\left(\lim_{n\to \infty}E_n\right) = \lim_{n\to \infty}\mu(E_n)$

  x. If $E_n \in \mathcal{R}$, , and $\exists k \in \mathbb{N}$ such that $\sum_{n=k}^{\infty}\mu(E_n) < \infty$, then

$\mu\left(\lim\sup_{n}E_n\right) = 0$