Almost-Everywhere Equality Relation for Real-Valued Functions is Equivalence Relation

Theorem

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

Let $\map {\mathcal M} {X, \Sigma, \R}$ be the real-valued $\Sigma$-measurable functions on $X$.

Let $\sim_\mu$ be the $\mu$-almost-everywhere equality relation on $\map {\mathcal M} {X, \Sigma, \R}$ by:

$f \sim_\mu g$ if and only if $\map f x = \map g x$ for $\mu$-almost all $x \in X$.

That is:

$\map \mu {\set {x \in X : \map f x \ne \map g x} } = 0$

Then $\sim_\mu$ is an equivalence relation.

Proof

Checking in turn each of the criteria for equivalence:

Reflexivity

From Equality is Reflexive, we have:

$\map f x = \map f x$ for $\mu$-almost all $x \in X$.

Therefore:

$f \sim_\mu f$

Hence $\sim_\mu$ is a reflexive relation.

$\Box$

Symmetry

Suppose:

$f \sim_\mu g$

Then:

$\map f x = \map g x$ for $\mu$-almost all $x \in X$.

Then by Equality is Symmetric:

$\map g x = \map f x$ for $\mu$-almost all $x \in X$.

Therefore:

$g \sim_\mu f$

Hence $\sim_\mu$ is a symmetric relation.

$\Box$

Transitivity

Suppose:

$f \sim_\mu g$

And:

$g \sim_\mu h$

Then:

$\map f x = \map g x$ for $\mu$-almost all $x \in X$.

And:

$\map g x = \map h x$ for $\mu$-almost all $x \in X$.

Then by Equality is Transitive:

$\map f x = \map h x$ for $\mu$-almost all $x \in X$.

Hence $\sim_\mu$ is a transitive relation.

$\Box$

All the criteria are therefore seen to hold for $\sim_\mu$ to be an equivalence relation.

$\blacksquare$