User:Caliburn/s/mt/Equality Almost Everywhere is Equivalence Relation/Lebesgue Space

Theorem

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

Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$.

Let $\sim_\mu$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^p} {X, \Sigma, \mu}$.

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

Proof

$\sim_\mu$ is Reflexive

We first show that $\sim_\mu$ is reflexive.

Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.

We have that:

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

from P-Seminorm of Function Zero iff A.E. Zero.

So $f \sim_\mu f$.

So $\mu_\sim$ is reflexive.

$\Box$

$\sim_\mu$ is Symmetric

We now show that $\sim_\mu$ is symmetric.

Let $f, g \in \map {\LL^p} {X, \Sigma, \mu}$.

Then:

$\norm {f - g}_p = \size {-1} \norm {g - f}_p = \norm {g - f}_p$

from property $(\text N 2)$ of the seminorm $\norm \cdot_p$.

So:

$\norm {f - g}_p = 0$ if and only if $\norm {g - f}_p = 0$

so:

$f \sim_\mu g$ if and only if $g \sim_\mu f$.

So $\sim_\mu$ is symmetric.

$\Box$

$\sim_\mu$ is Transitive

We finally show that $\sim_\mu$ is transitive.

Let $f, g, h \in \map {\LL^p} {X, \Sigma, \mu}$ be such that $f \sim_\mu g$ and $g \sim_\mu h$.

Then:

$\norm {f - g}_p = 0$ and $\norm {g - h}_p = 0$.

Then:

\(\ds \norm {f - h}_p\)

\(=\)

\(\ds \norm {\paren {f - g} - \paren {g - h} }_p\)

\(\ds \)

\(\le\)

\(\ds \norm {f - g}_p + \norm {g - h}_p\)

Minkowski's Inequality on Lebesgue Space

\(\ds \)

\(=\)

\(\ds 0\)