Lp Norm is Well-Defined

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

Let $\map {\mathcal M} {X, \Sigma}$ be the set of $\Sigma$-measurable functions on $X$.

Let $\sim$ be the almost-everywhere equality equivalence relation on $\map {\mathcal M} {X, \Sigma}$.

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

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

Then the $L^p$ norm:


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

is well-defined.

Proof
Note that the integral 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:


 * $\norm f_p = \norm g_p$

From Equivalence Class Equivalent Statements, we have:


 * $f \sim g$

So, from the definition of the almost-everywhere equality equivalence 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 almost all $x \in X$.

So, from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 2, we have:


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

That is:


 * $\norm f_p = \norm g_p$

as required.