User:Caliburn/s/mt/Equality Almost Everywhere is Equivalence Relation/Measurable Functions
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma}$ be the space of $\Sigma$-measurable functions on $\struct {X, \Sigma}$.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the space of real-valued $\Sigma$-measurable functions on $\struct {X, \Sigma}$.
Let $\mathcal S \in \set {\map {\mathcal M} {X, \Sigma}, \map {\mathcal M} {X, \Sigma, \R} }$
Let $\sim_\mu$ be the $\mu$-almost-everywhere equality relation on $\mathcal S$.
Then $\sim_\mu$ is an equivalence relation.
Proof
$\sim_\mu$ is Reflexive
We first show that $\sim_\mu$ is reflexive.
Let $f \in \mathcal S$.
Note that we have:
- $\set {x \in X : \map f x \ne \map f x} = \O$
so, from Empty Set is Null Set:
- $\map \mu {\set {x \in X : \map f x \ne \map f x} } = 0$
so:
- $f \sim_\mu f$
So $\sim_\mu$ is reflexive.
$\Box$
$\sim_\mu$ is Symmetric
We now show that $\sim_\mu$ is symmetric.
Let $f, g \in \mathcal S$ be such that $f \sim_\mu g$.
Then:
- $\map \mu {\set {x \in X : \map f x \ne \map g x} } = 0$
We have:
- $\set {x \in X : \map f x \ne \map g x} = \set {x \in X : \map g x \ne \map f x}$
so:
- $\map \mu {\set {x \in X : \map g x \ne \map f x} } = 0$
hence:
- $g \sim_\mu f$
So if $f \sim_\mu g$ then $g \sim_\mu f$.
So $\sim_\mu$ is symmetric.
$\Box$
$\sim_\mu$ is Transitive
We finally show that $\sim_\mu$ is transitive.
Let $f, g, h \in \mathcal S$ be such that $f \sim_\mu g$ and $g \sim_\mu h$.
Then, we have:
- $\map \mu {\set {x \in X : \map f x \ne \map g x} } = 0$
and:
- $\map \mu {\set {x \in X : \map g x \ne \map h x} } = 0$
From Null Sets Closed under Countable Union, we have:
- $\map \mu {\set {x \in X : \map f x \ne \map g x} \cup \set {x \in X : \map g x \ne \map h x} } = 0$
Now let $x \in X$ be such that $\map f x \ne \map h x$.
Then we cannot have both $\map f x = \map g x$ and $\map g x = \map h x$, otherwise $\map f x = \map h x$ from Equality is Transitive.
So, we have $x \in \set {x \in X : \map f x \ne \map g x} \cup \set {x \in X : \map g x \ne \map h x}$.
So:
- $\set {x \in X : \map f x \ne \map h x} \subseteq \set {x \in X : \map f x \ne \map g x} \cup \set {x \in X : \map g x \ne \map h x}$
So:
- $\map \mu {\set {x \in X : \map f x \ne \map h x} } = 0$
from Null Sets Closed under Subset.
So $f \sim_\mu h$.
So whenever $f \sim_\mu g$ and $g \sim_\mu h$ we have $f \sim_\mu h$.
So $\sim_\mu$ is transitive.