Lp Norm is Well-Defined

Theorem

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

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

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

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

Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$.

$\ds \norm {\eqclass f \sim}_p = \paren {\norm f_p}^{1/p}$

is well-defined.

Note that:

$\ds \int \size f^p \rd \mu$

is well-defined from the definition of the Lebesgue $p$-space.

We show that for $E \in \map {L^p} {X, \Sigma, \mu}$, $\norm E_p$ is independent of the representative chosen for $E$.

Let:

$E = \eqclass f \sim = \eqclass g \sim$

for $\eqclass f \sim, \eqclass g \sim \in \map {L^p} {X, \Sigma, \mu}$.

We show that:

$\norm f_p = \norm g_p$

$f \sim g$

So, from the definition of the almost-everywhere equality relation, we have:

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

That is, there exists a $\mu$-null set $N$ such that:

if $\map f x \ne \map g x$ then $x \in N$.

If $\size {\map f x}^p \ne \size {\map g x}^p$, we have $\map f x \ne \map g x$, so:

if $\size {\map f x}^p \ne \size {\map g x}^p$ then $x \in N$.

So:

$\size {\map f x}^p = \size {\map g x}^p$ for $\mu$-almost all $x \in X$.

$\ds \paren {\int \size f^p \rd \mu}^{1/p} = \paren {\int \size g^p \rd \mu}^{1/p}$

That is:

$\norm f_p = \norm g_p$

as required.

$\blacksquare$