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.

$\sim_\mu$ 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.

$\sim_\mu$ 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.

$\sim_\mu$ 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.