User:Caliburn/s/mt/Relation Identifying A.E. Equal Functions is Equivalence Relation

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\R^X$ be the set of real-valued functions $f : X \to \R$.

Define a relation $\RR$ on $\R^X$ by:


 * $f \RR g$ $f = g$ $\mu$-almost everywhere.

Then $\RR$ is an equivalence relation on $\R^X$.

Proof
We verify each of the three conditions of a equivalence relation.

Reflexivity
Note that if $f = f$, we have:


 * $\map f x = \map f x$ for all $x \in X$.

We can write this:


 * if $x \in X$ has $\map f x \ne \map f x$ then $x \in \O$

which is vacuously true.

From Empty Set is Null Set, we have:


 * $\map \mu \O = 0$

So:


 * $f = f$ $\mu$-almost everywhere.

So:


 * $f \RR f$

So $\RR$ is a reflexive relation.

Symmetry
Suppose that $f, g \in \R^X$ have:


 * $f \RR g$

Then:


 * $f = g$ $\mu$-almost everywhere.

So there exists a $\mu$-null set $N \subseteq X$ such that:


 * whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$.

From Equality is Symmetric, we have:


 * whenever $x \in X$ has $\map g x \ne \map f x$, we have $x \in N$.

so:


 * $g = f$ $\mu$-almost everywhere.

So:


 * $f \RR g$ implies $g \RR f$.

So $\RR$ is a symmetric relation.

Transitive
Suppose that:


 * $f \RR g$

and:


 * $g \RR h$

Then:


 * $f = g$ $\mu$-almost everywhere

and:


 * $g = h$ $\mu$-almost everywhere.

So there exists a $\mu$-null set $N_1 \subseteq X$ such that:


 * whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N_1$

and there exists a $\mu$-null set $N_2 \subseteq X$ such that:


 * whenever $x \in X$ has $\map g x \ne \map h x$, we have $x \in N_2$.

If $x \in X$ was such that:


 * $\map f x = \map g x$

and:


 * $\map g x = \map h x$

then we would certainly have:


 * $\map f x = \map h x$

from Equality is Transitive.

So, if $x \in X$ is such that:


 * $\map f x \ne \map h x$

then either:


 * $\map f x \ne \map g x$

or:


 * $\map g x \ne \map h x$

That is:


 * $x \in N_1$ or $x \in N_2$

From the definition of set union, we have:


 * $x \in N_1 \cup N_2$

From Null Sets Closed under Countable Union, we have:


 * $N_1 \cup N_2$ is $\mu$-null.

So:


 * if $x \in X$ is such that $\map f x \ne \map h x$, then $x \in N_1 \cup N_2$, a $\mu$-null set.

So:


 * $f = h$ $\mu$-almost everywhere

giving:


 * $f \RR g$ and $g \RR h$ implies $f \RR h$.

So $\RR$ is a transitive relation.

Since $\RR$ is reflexive, symmetric and transitive, it is an equivalence relation.