Almost-Everywhere Equality Relation for Lebesgue Space is Equivalence Relation/Proof 1

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.

Let $\map {\mathcal L^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space of $\struct {X, \Sigma, \mu}$.

We define the $\mu$-almost-everywhere equality relation $\sim_\mu$ on $\map {\mathcal L^p} {X, \Sigma, \mu}$ by:

$\forall f, g \in \mathcal L^p: f \sim_\mu g$ if and only if $\norm {f - g}_p = 0$

where $\norm {\, \cdot \,}_p$ is the $p$-seminorm.

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


Proof

Let $f, g, h \in \LL^p$.

Checking in turn each of the criteria for equivalence:


Reflexivity

From P-Seminorm of Function Zero iff A.E. Zero, we have:

$\norm {f - f}_p = 0$

Therefore:

$f \sim_\mu f$

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

$\Box$


Symmetry

Suppose:

$f \sim_\mu g$

Then:

$\norm {f - g}_p = 0$

Then:

$\norm {g - f}_p = 0$

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:

$\norm {f - g}_p = 0$

and:

$\norm {g - h}_p = 0$

Then:

$\norm {f - h}_p = 0$

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

$\Box$


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

$\blacksquare$


Sources


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