Sigma Algebra, Measure and Normed Ring

Let $(X,\mathrm{\Sigma },\mu )$$(X,\Sigma,\mu)$ be a measurable space. We require that $\mu <\mathrm{\infty }$$\mu < \infty$.

We know that the sigma algebra $(\mathrm{\Sigma },\mathrm{\Delta },\cap )$$(\Sigma,\Delta,\cap)$ is a Boolean ring, hence a ${\mathbb{F}}_2$$\mathbb{F}_2$ algebra. Here $A\mathrm{\Delta }B:=(A\cup B)-(A\cap B)$$A \Delta B := (A \cup B) - (A \cap B)$​​ is the symmetric difference.

Easy to see that $(\mathrm{\Sigma },d)$$(\Sigma,d)$ with $d(A,B)=\mu (A\mathrm{\Delta }B)$$d(A,B)=\mu(A\Delta B)$ is a pseudometric.

Proposition.

Proposition. The "kernel" of $\mu$$\mu$, $I=\{A\in \mathrm{\Sigma }:\mu (A)=0\}$$I = \{A \in \Sigma: \mu(A) = 0\}$, is an ideal of $(\mathrm{\Sigma },\mathrm{\Delta },\cap )$$(\Sigma,\Delta,\cap)$.

Proof. $\mathrm{\forall }A,B\in I,0\le \mu (A\mathrm{\Delta }B)\le \mu (A)+\mu (B)=0,0\le \mu (A\cap B)\le \mu (A)=0.$$\forall A,B \in I, 0 \le \mu(A \Delta B) \le \mu(A) + \mu(B) = 0, 0 \le \mu(A \cap B) \le \mu(A) = 0.$ $◻$$\square$.

Let us consider the quotient ring $\pi :\mathrm{\Sigma }\to \mathrm{\Sigma }/I$$\pi: \Sigma \to \Sigma/I$, and define ${\mu }^{\prime }:\mathrm{\Sigma }/I\to {\mathbb{R}}^{+}\cup \{\mathrm{\infty }\}$$\mu': \Sigma/I \to \mathbb{R}^{+} \cup \{\infty\}$ to be ${\mu }^{\prime }(\pi [A])=\mu (A)$$\mu'(\pi[A]) = \mu(A)$.

This is well-defined, since if $\pi (A)=\pi (B)$$\pi(A) = \pi(B)$, then $A=B\mathrm{\Delta }C,C\in I$$A = B \Delta C, C \in I$. This implies $A\mathrm{\Delta }C=(B\mathrm{\Delta }C)\mathrm{\Delta }C=B$$A \Delta C = (B \Delta C) \Delta C = B$, and

$$\begin{array}{}\text{(1)}& \mu (A)=\mu (B\mathrm{\Delta }C)\le \mu (B)+\mu (C)=\mu (B),\end{array}$$

and

$$\begin{array}{}\text{(2)}& \mu (B)=\mu (A\mathrm{\Delta }C)\le \mu (A)+\mu (C)=\mu (A).\end{array}$$

Hence $\mu (A)\le \mu (B)\le \mu (A)$$\mu(A) \le \mu(B) \le \mu(A)$, thus $\mu (A)=\mu (B)$$\mu(A) = \mu(B)$.

Proposition. The function $d(\pi [A],\pi [B]):={\mu }^{\prime }(\pi [A\mathrm{\Delta }B])$$d(\pi[A], \pi[B]) := \mu'(\pi[A \Delta B])$ is a metric on $\mathrm{\Sigma }/I$$\Sigma/I$.

Proof. Since $0\le \mu <\mathrm{\infty }$$0 \le \mu < \infty$, then $0\le {\mu }^{\prime }<\mathrm{\infty }$$0 \le \mu' < \infty$.

${\mu }^{\prime }(\pi [A])=\mu (A)=0⟺A\in I$$\mu'(\pi[A]) = \mu(A) = 0 \iff A \in I$ by definition of $I$$I$,

hence $d(\pi [A],\pi [B])=0⟺{\mu }^{\prime }(\pi [A\mathrm{\Delta }B])=0⟺\mu (A\mathrm{\Delta }B)=0⟺A\mathrm{\Delta }B\in I⟺\pi [A]=\pi [B]$$d(\pi[A], \pi[B]) = 0 \iff \mu'(\pi[A \Delta B]) = 0 \iff \mu(A \Delta B) = 0 \iff A \Delta B \in I \iff \pi[A] = \pi[B]$.

It is easy to see that $d$$d$ is symmetric.

For the triangle inequality, ${\mu }^{\prime }(\pi [A\mathrm{\Delta }C])=\mu ((A\mathrm{\Delta }B)\mathrm{\Delta }(B\mathrm{\Delta }C))\le \mu (A\mathrm{\Delta }B)+\mu (B\mathrm{\Delta }C)={\mu }^{\prime }(\pi [A\mathrm{\Delta }B])+{\mu }^{\prime }(\pi [B\mathrm{\Delta }C])$$\mu'(\pi[A \Delta C]) = \mu((A \Delta B) \Delta (B \Delta C)) \le \mu(A \Delta B) + \mu(B \Delta C) = \mu'(\pi[A \Delta B]) + \mu'(\pi[B \Delta C])$. $◻$$\square$

Hence we get a metric space $(\mathrm{\Sigma }/I,d)$$(\Sigma/I, d)$. If the "kernel" of $\mu$$\mu$ is $\{\mathrm{\varnothing }\}$$\{\emptyset\}$, then what we get is $(\mathrm{\Sigma },d)$$(\Sigma, d)$.

For example, let $X$$X$ be a finite set and $\mathrm{\Sigma }:=2^X,\mu (A):=|A|$$\Sigma := 2^X, \mu(A) := |A|$. Define $d(A,B):=|A\mathrm{\Delta }B|$$d(A,B) := |A \Delta B|$.

Proposition. $(2^X,|\cdot |)$$(2^X, |\cdot|)$ is a normed ring.

Proof. Obviously $|A|=0⟺A=\mathrm{\varnothing }$$|A| = 0 \iff A = \emptyset$, $-A=A⟹|-A|=|A|,|A\mathrm{\Delta }B|\le |A|+|B|,|A\cap B|\le |A||B|.$$-A = A \implies |-A| = |A|, |A \Delta B| \le |A| + |B|, |A \cap B| \le |A||B|.$

We could generalize it as follows.

Proposition. $(\mathrm{\Sigma },\mu )$$(\Sigma, \mu)$ is a normed ring iff $\mathrm{\forall }A\ne \mathrm{\varnothing },1\le \mu (A)<\mathrm{\infty }$$\forall A \neq \emptyset, 1\le\mu(A) <\infty$.

Proof.

If $\mathrm{\forall }A\ne \mathrm{\varnothing },\mu (A)\ge 1$$\forall A \neq \emptyset, \mu(A) \geq 1$.${\mu }^{-1}(0)=\{\mathrm{\varnothing }\}$$\mu^{-1}(0)=\{\empty\}$, so $\mu (A)=0⟺A=\mathrm{\varnothing }$$\mu(A)=0\iff A=\empty$.

Then we only need to check that $|fg|\le |f||g|$$|fg| \le |f||g|$.

If $\mathrm{\forall }A\ne \mathrm{\varnothing },\mu (A)\ge 1$$\forall A \neq \emptyset, \mu(A) \geq 1$, then $\mu (A\cap B)\le \mu (A)\le \mu (A)\mu (B)$$\mu(A \cap B) \le \mu(A) \le \mu(A)\mu(B)$.

If there exists a set $A\ne \mathrm{\varnothing }$$A \neq \emptyset$ such that $\mu (A)<1$$\mu(A) < 1$, then $\mu (A)=\mu (A\cap A)>\mu (A)\mu (A)$$\mu(A) = \mu(A \cap A) > \mu(A)\mu(A)$, hence it is not a normed algebra. $◻$$\square$

Corollary. If $inf\{\mu (A):A\in \mathrm{\Sigma }-\{\mathrm{\varnothing }\}\}$$\inf\{\mu(A): A \in \Sigma - \{\emptyset\}\}$ exists and not equal to zero, denote it as ${\alpha }^{-1}$$\alpha^{-1}$, then $(\mathrm{\Sigma },\alpha \mu )$$(\Sigma, \alpha\mu)$ is a normed algebra.

There are still some problems I need to think about, such as the relationships between completeness, connectedness, compactness, and the measurable space.