Category of Measurable space and relative functor

Category of measurable spaceSigma Algebra and measurable spaceInternal hom in sigma algebraMeasurable functionBorel funcorLeft and Right adjoint of forgetful functor MeasureA functor from Meas to MonProbability SpaceSubobject Conditional probability

 

Category of measurable space

Sigma Algebra and measurable space

Definition. A collection $\mathrm{\Omega }$$\Omega$ of subsets of a set $M$$M$ is said to be a $\sigma -$$\sigma-$algebra in $M$$M$ if $\mathrm{\Omega }$$\Omega$ has following properties:

• $M\in \mathrm{\Omega }$$M\in\Omega$

• If $A\in \mathrm{\Omega }$$A\in\Omega$ then $A^c\in \mathrm{\Omega }$$A^c\in\Omega$. Where $A^c$$A^c$ is the complement of $A$$A$ relative to $M$$M$.

• If ${\displaystyle A=\bigcup _{i=1}^{\mathrm{\infty }}{A}_i}$$\displaystyle A=\bigcup_{i=1}^\infty A_i$ if $A_i\in \mathrm{\Omega }$$A_i\in\Omega$ for $i\in \mathbb{N}$$i\in\mathbb N$ then $A\in \mathrm{\Omega }$$A\in\Omega$. In other word, $\mathrm{\Omega }$$\Omega$ is closed under countable union.

A pair $(M,\mathrm{\Omega })$$(M,\Omega)$​ is called measurable space if $M$$M$ is a set and $\mathrm{\Omega }$$\Omega$ is a sigma algebra over $M$$M$.

Remark.

Readers could think about the analogy between category of measurable space and Category of topology space.

Proposition.

• $\mathrm{\varnothing }\in \mathrm{\Omega }$$\empty\in\Omega$

• If ${\displaystyle A=\bigcap _{i=1}^{\mathrm{\infty }}{A}_i}$$\displaystyle A=\bigcap_{i=1}^\infty A_i$ if $A_i\in \mathrm{\Omega }$$A_i\in\Omega$ for $i\in \mathbb{N}$$i\in\mathbb N$ then $A\in \mathrm{\Omega }$$A\in\Omega$. In other word, $\mathrm{\Omega }$$\Omega$​ is closed under countable intersection

• $A,B\in \mathrm{\Omega }$$A,B\in\Omega$ implies $A-B\in \mathrm{\Omega }$$A-B\in\Omega$.

Proof.

• $M\in \mathrm{\Omega }$$M\in\Omega$ and $M^c=\mathrm{\varnothing }$$M^c=\empty$, hence $\mathrm{\varnothing }\in \mathrm{\Omega }$$\empty\in\Omega$​.

• ${\displaystyle \bigcup _{i=1}^{\mathrm{\infty }}{A}_i^c={\left(\bigcap _{i=1}^{\mathrm{\infty }}{A}_i\right)}^c\in \mathrm{\Omega }}$$\displaystyle\bigcup_{i=1}^{\infty} A_i^c=\left(\bigcap_{i=1}^{\infty}A_i\right)^c\in\Omega$, hence ${\displaystyle \bigcap _{i=1}^{\mathrm{\infty }}{A}_i\in \mathrm{\Omega }}$$\displaystyle\bigcap_{i=1}^{\infty}A_i\in\Omega$.

• $A-B=A\cap B^c.◻$$A-B=A\cap B^c.\quad\square$

Hence sigma algebra is a kind of Boolean Algebra! We will meet $\mathrm{B}\mathrm{o}\mathrm{o}\mathrm{l}$$\rm Bool$ later, when we define measurable function.

Internal hom in sigma algebra

Let $(M,\mathrm{\Omega })$$(M,\Omega)$ be a measurable space.

Sometime, in particular in probability theory, For $E_1,E_2\in \mathrm{\Omega }$$E_1,E_2\in\Omega$, we say $E_1⟹E_2⟺E_1\subseteq E_2$$E_1\implies E_2\iff E_1\subseteq E_2$.

Remember that in logic, $p⟹q$$p\implies q$ is the internal hom, and we have the tensor-hom adjoint as follows:

$$\begin{array}{}\text{(1)}& p\wedge q\to r⟺p\to (q⟹r)\end{array}$$

Here $q⟹r$$q\implies r$ is a proposition as well.

We would like to do the same things for $\mathrm{\Omega }$$\Omega$, luckily, it is a Boolean Algebra, hence we could define

$$\begin{array}{}\text{(2)}& E_1⟹E_2:=E_1^c\cup E_2\end{array}$$

Proposition. $E_1\subseteq E_2⟺E_1⟹E_2=M$$E_1\subseteq E_2\iff E_1\implies E_2=M$.

Proof.

If $E_1\subseteq E_2,$$E_1\subseteq E_2,$ then $E_1^c\supseteq E_2^c$$E_1^c\supseteq E_2^c$. Hence $M\supseteq E_1^c\cup E_2\supseteq E_2^c\cup E_2=M$$M\supseteq E_1^c\cup E_2\supseteq E_2^c\cup E_2=M$​.

If $E_1⟹E_2=M$$E_1\implies E_2=M$, then $E_1^c\cup E_2=M⟹E_1\cap E_2^c=\mathrm{\varnothing }$$E_1^c\cup E_2=M\implies E_1\cap E_2^c=\empty$. Hence $E_1\subseteq E_2$$E_1\subseteq E_2$. $◻$$\square$

Proposition. $E_1⟹E_2=\mathrm{\varnothing }⟺E_1=M\wedge E_2=\mathrm{\varnothing }$$E_1\implies E_2 =\empty\iff E_1=M\wedge E_2=\empty$.

Proof. Obviously.

Measurable function

Definition.

Let $(X,\mathrm{\Sigma }),(Y,\mathrm{\Omega })$$(X,\Sigma),(Y,\Omega)$ be to measurable space. A function $f:X\to Y$$f:X\to Y$ is measurable if $f^{-1}:\mathrm{\Omega }\to \mathrm{\Sigma }$$f^{-1}:\Omega\to\Sigma$.

i.e. if $V\in \mathrm{\Omega }$$V\in\Omega$, then $f^{-1}(V)\in \mathrm{\Sigma }$$f^{-1}(V)\in\Sigma$​.

Definition. The object of category of measurable space $\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{s}$$\rm Meas$ is measurable space and morphism in $\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{s}$$\rm Meas$ is measurable functions.

The definition of measurable function tells us that there is a functor $D:\mathrm{M}\mathrm{e}\mathrm{a}{\mathrm{s}}^{\mathrm{op}}\to \mathrm{B}\mathrm{o}\mathrm{o}\mathrm{l}$$D:\rm Meas^{\mathrm{op}}\to Bool$.

$$\begin{array}{}\text{(3)}& D(X,\mathrm{\Sigma })=\mathrm{\Sigma },D(f)=f^{-1}\end{array}$$

Borel funcor

Definition. A Borel set is any set in a topological space that can be formed from open sets or closed sets through countable union, countable intersection, and relative complement. For a topological space $(X,\tau )$$(X,\tau)$, we can use Borel set to get a measurable space $(X,\mathcal{B}(\tau ))$$(X,\mathcal B(\tau))$. Indeed, Borel set gives us a functor from $\mathrm{Top}\to \mathrm{Meas}$$\mathrm{Top}\to\mathrm{Meas}$ as follows.

$$\begin{array}{}\text{(4)}& \mathcal{B}:\mathrm{Top}\to \mathrm{Meas},(X,\tau )⟼(X,\mathcal{B}(\tau )),f:(X,\tau )\to (Y,{\tau }^{\prime })⟼f:(X,\mathcal{B}(\tau ))\to (Y,\mathcal{B}({\tau }^{\prime }))\end{array}$$

Easy to see that $f:(X,\mathcal{B}(\tau ))\to (Y,\mathcal{B}({\tau }^{\prime }))$$f:(X,\mathcal B(\tau))\to(Y,\mathcal B(\tau'))$ is a measurable functor as well.

It also cam be viewed as a functor from Heyting Algebra to Boolean Algebra.

Left and Right adjoint of forgetful functor

Let $\mathcal{F}:\mathrm{Meas}\to \mathrm{Set}$$\mathcal F:\mathrm{Meas}\to\mathrm{Set}$ be the forgetful functor, Then easy to see that

$$\begin{array}{}\text{(5)}& \mathcal{D}:X⟼(X,P(X)),f:X\to Y⟼f:(X,P(X))\to (Y,P(Y))\end{array}$$

is the left adjoint of $\mathcal{F}$$\mathcal F$. Since

$$\begin{array}{}\text{(6)}& {\mathrm{Hom}}_{\mathrm{Meas}}(\mathcal{D}(A),B)\cong {\mathrm{Hom}}_{\mathrm{Set}}(A,\mathcal{F}(B))\end{array}$$

Similarly,

$$\begin{array}{}\text{(7)}& \mathcal{T}:X⟼(X,\{\mathrm{\varnothing },X\}),f:X\to Y⟼f:(X,\{\mathrm{\varnothing },X\})\to (Y,\{\mathrm{\varnothing },Y\})\end{array}$$

is the right adjoint of $\mathcal{F}$$\mathcal F$.

Readers should compare it with Math Essays: Discrte Topology and Trivial topology: An adjoint functor point of view. (marco-yuze-zheng.blogspot.com).

Measure

Let ${\displaystyle (X,\mathrm{\Sigma })}$$\displaystyle(X,\Sigma)$ be a measurable space, a function $\mu :\mathrm{\Sigma }\to {\mathbb{R}}^{+}\cup \{\mathrm{\infty }\}$$\mu:\Sigma\to\mathbb R^+\cup\{\infty\}$ is called measure if the following conditions hold:

$$\begin{array}{}\text{(8)}& \mu (\mathrm{\varnothing })=0\end{array}$$

For all countable collections ${\displaystyle \{E_i{\}}_{i=1}^{\mathrm{\infty }}}$$\displaystyle\{E_i\}_{i=1}^{\infty}$ of point wise disjoint sets in $\mathrm{\Sigma }$$\Sigma$,

$$\begin{array}{}\text{(9)}& {\displaystyle \mu \left(\bigcup _{i=1}^{\mathrm{\infty }}{E}_k\right)=\sum _{i=1}^{\mathrm{\infty }}\mu (E_i).}\end{array}$$

Remark. If there exists a $E\in \mathrm{\Sigma }$$E\in\Sigma$ such that $\mu (E)<\mathrm{\infty }$$\mu(E)<\infty$, then automatically $\mu (\mathrm{\varnothing })=0$$\mu(\empty)=0$ since $\mu (E\cup \mathrm{\varnothing })=\mu (E)+\mu (\mathrm{\varnothing })$$\mu(E\cup\empty)=\mu(E)+\mu(\empty)$.

If we consider $\mathcal{T}({\mathbb{R}}^{+}\cup \{\mathrm{\infty }\})\in \mathrm{Ob}(\mathrm{Meas})$$\mathcal T(\mathbb R^+\cup\{\infty\})\in\mathrm{Ob}(\mathrm{Meas})$, then every function will be measurable.

Therefore, $\mu :(X,\mathrm{\Sigma })\to \mathcal{T}({\mathbb{R}}^{+}\cup \{\mathrm{\infty }\})$$\mu:(X,\Sigma)\to\mathcal T(\mathbb R^+\cup\{\infty\})$​ is measurable function.

Let $f:(X,\mathrm{\Sigma })\to (Y,\mathrm{\Omega })$$f:(X,\Sigma)\to(Y,\Omega)$ be a measurable function and $\mu$$\mu$ be a measure on $(X,\mathrm{\Sigma })$$(X,\Sigma)$, we can induce a measure on $Y$$Y$ via $f$$f$

by consider ${\mu }^{\prime }:=\mu \circ f^{-1}$$\mu':=\mu\circ f^{-1}$. Easy to see this form a measure as well.

A functor from Meas to Mon

Let us denote the set of measure on $(X,\mathrm{\Sigma })$$(X,\Sigma)$ as $\mathcal{M}(X,\mathrm{\Sigma })$$\mathcal M(X,\Sigma)$. This give us a functor to $\mathrm{AbMon}$$\mathrm{AbMon}$ by

$$\begin{array}{}\text{(10)}& \mathcal{M}:(X,\mathrm{\Sigma })\to \mathcal{M}(X,\mathrm{\Sigma })\end{array}$$

Which is a monoid. Let $\mu ,{\mu }^{\prime }\in \mathcal{M}(X,\mathrm{\Sigma })$$\mu,\mu'\in\mathcal M(X,\Sigma)$, then $\mu +{\mu }^{\prime }\in \mathcal{M}(X,\mathrm{\Sigma })$$\mu+\mu'\in\mathcal M(X,\Sigma)$​ since

$$\begin{array}{}\text{(11)}& \mu \ge 0,{\mu }^{\prime }\ge 0⟹\mu +{\mu }^{\prime }\ge 0\end{array}$$

$$\begin{array}{}\text{(12)}& (\mu +{\mu }^{\prime })(\mathrm{\varnothing })=\mu (\mathrm{\varnothing })+{\mu }^{\prime }(\mathrm{\varnothing })=0+0=0\end{array}$$

Also for countable collections ${\displaystyle \{E_i{\}}_{i=1}^{\mathrm{\infty }}}$$\displaystyle\{E_i\}_{i=1}^{\infty}$ of point wise disjoint sets in $\mathrm{\Sigma }$$\Sigma$

$$\begin{array}{}\text{(13)}& (\mu +{\mu }^{\prime })\left(\bigcup _{i=1}^{\mathrm{\infty }}{E}_i\right)=\mu \left(\bigcup _{i=1}^{\mathrm{\infty }}{E}_i\right)+{\mu }^{\prime }\left(\bigcup _{i=1}^{\mathrm{\infty }}{E}_i\right)=\sum _{i=1}^{\mathrm{\infty }}\mu (E_i)+\sum _{i=1}^{\mathrm{\infty }}{\mu }^{\prime }(E_i)=\sum _{i=1}^{\mathrm{\infty }}(\mu +{\mu }^{\prime })(E_i)\end{array}$$

For morphism, let $f:(X,\mathrm{\Sigma })\to (Y,\mathrm{\Omega })$$f:(X,\Sigma)\to (Y,\Omega)$​ be a measureable function

$$\begin{array}{}\text{(14)}& \mathcal{M}(f):\mathcal{M}(X,\mathrm{\Sigma })\to \mathcal{M}(Y,\mathrm{\Omega }),\mu ⟼\mu \circ f^{-1}\end{array}$$

Also

$$\begin{array}{}\text{(15)}& \mathcal{M}(f)(\mu +{\mu }^{\prime })=(\mu +{\mu }^{\prime })\circ f^{-1}=\mu \circ f^{-1}+{\mu }^{\prime }\circ f^{-1}=\mathcal{M}(f)(\mu )+\mathcal{M}(f)({\mu }^{\prime })\end{array}$$

Finally

$$\begin{array}{}\text{(16)}& \mathcal{M}(f\circ g)(\mu )=\mu \circ (f\circ g{)}^{-1}=\mu \circ g^{-1}\circ f^{-1}\end{array}$$

$$\begin{array}{}\text{(17)}& \mathcal{M}(f)\circ \mathcal{M}(g)(\mu )=(\mu \circ g^{-1})\circ f^{-1}=\mu \circ g^{-1}\circ f^{-1}\end{array}$$

Hence $\mathcal{M}$$\mathcal M$ is a functor. $◻$$\square$

Probability Space

A probability space is a measurable space $(E,\mathrm{\Sigma },P)$$(E,\Sigma,P)$. Where $E$$E$ is the underline set, $\mathrm{\Sigma }$$\Sigma$ is the sigma algebra, and $P$$P$ is the probability measure, whcih is a measure satisfy $P(E)=1$$P(E)=1$​.

Subobject

Let $(M,\mathrm{\Sigma })$$(M,\Sigma)$ be a measurable space, a subobject of $(M,\mathrm{\Sigma })$$(M,\Sigma)$ is a measurable space $(N,\mathrm{\Omega })$$(N,\Omega)$ with a monomorphism

$$\begin{array}{}\text{(18)}& \iota :(N,\mathrm{\Omega })\to (M,\mathrm{\Sigma })\end{array}$$

Notice that this form a pre-order set, and we could do the posetlization for it. i.e. quotient the isomorphism relation.

So usually we choose the subset of $M$$M$ to be the representative element of the equivalent class.

Define the sub measurable space on $N\subseteq M$$N\subseteq M$ to be the greatest (in the poset of subobject) measurable space $(N,\mathrm{\Omega })$$(N,\Omega)$ on $N$$N$.

Conditional probability

Let $(E,\mathrm{\Sigma },P)$$(E,\Sigma,P)$ be a probability space, $E_1\in \mathrm{\Sigma }$$E_1\in\Sigma$ is an measurable set, we call it event. We can consider the sub measurable space on $E_1$$E_1$ and induce a new probability measure, ${\displaystyle (E_1,\mathrm{\Omega },P(-|E_1)}$$\displaystyle\left(E_1,\Omega,P(-|E_1\right)$, where $P(-|E_1):=\frac{P(-\cap E_1)}{P(E_1)}$$P(-|E_1):=\frac{P(-\cap E_1)}{P(E_1)}$.

This is the conditional probability.

Probability Reciprocity and Quadratic Reciprocity

$$\begin{array}{}\text{(19)}& P(E_1\cap E_2)=P(E_1|E_2)P(E_2)=P(E_2\cap E_1)=P(E_2|E_1)P(E_1)\end{array}$$

Let us define $(E_i):=P(E),P(E_i|E_j):=(E_i|E_j)$$(E_i):=P(E),P(E_i|E_j):=(E_i|E_j)$.

According to $(19)$$(19)$, we have

$$\begin{array}{}\text{(20)}& (E_1|E_2)(E_2)=(E_2|E_1)(E_1),⟹(E_1|E_2)=(E_2|E_1)\frac{(E_1)}{(E_2)}\text{If }(E_2)\ne 0\end{array}$$

Hence

$$\begin{array}{}\text{(21)}& (E_1|E_2)=(E_2|E_1)⟺(E_1)=(E_2)\end{array}$$

Also, reader should compare it with Quadratic Reciprocity.

$$\begin{array}{}\text{(22)}& (E_1|E_2)=(E_2|E_1)\frac{(E_1)}{(E_2)}\end{array}$$

 

$$\begin{array}{}\text{(23)}& \left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)(-1{)}^{\frac{p-1}{2}\cdot \frac{q-1}{2}}\end{array}$$