L-Infinity Norm is Well-Defined
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space for $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ space on $\struct {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
Then the $L^\infty$ norm:
- $\ds \norm {\eqclass f \sim}_\infty = \norm f_\infty$
is well-defined.
Proof
We show that for $E \in \map {L^\infty} {X, \Sigma, \mu}$, $\norm E_\infty$ is independent of the representative chosen for $E$.
Let:
- $E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
We show that:
- $\norm f_\infty = \norm g_\infty$
From Equivalence Class Equivalent Statements, we have:
- $f \sim g$
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 $x \in N$.
We show that:
- $\map \mu {\set {x \in X : \size {\map f x} \ge c} } = \map \mu {\set {x \in X : \size {\map g x} \ge c} }$
From this it will follow that:
- $\inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0} = \inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map g x} \ge c} } = 0}$
That is:
- $\norm f_\infty = \norm g_\infty$
Clearly:
- $\set {x \in N : \size {\map f x} \ge c} \subseteq N$
So, from Null Sets Closed under Subset:
- $\map \mu {\set {x \in N : \size {\map f x} \ge c} } = 0$
Swapping $f$ for $g$, we also obtain:
- $\map \mu {\set {x \in N : \size {\map g x} \ge c} } = 0$
Since:
- $\set {x \in N : \size {\map f x} \ge c} \subseteq N$
and:
- $\set {x \in X \setminus N : \size {\map f x} \ge c} \subseteq X \setminus N$
we have that:
- $\set {x \in N : \size {\map f x} \ge c}$ and $\set {x \in X \setminus N : \size {\map f x} \ge c}$ are disjoint for all real numbers $c > 0$.
Then:
\(\ds \map \mu {\set {x \in X : \size {\map f x} \ge c} }\) | \(=\) | \(\ds \map \mu {\set {x \in N : \size {\map f x} \ge c} \cup \set {x \in X \setminus N : \size {\map f x} \ge c} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in N : \size {\map f x} \ge c} } + \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge c} }\) | from the countable additivity of $\mu$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge c} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge c} }\) | since $\map f x = \map g x$ for all $x \in X \setminus N$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in N : \size {\map g x} \ge c} } + \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge c} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in N : \size {\map g x} \ge c} \cup \set {x \in X \setminus N : \size {\map g x} \ge c} }\) | using the countable additivity of $\mu$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in X : \size {\map g x} \ge c} }\) |
as required.
$\blacksquare$