L-Infinity Norm is Well-Defined

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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.

$\blacksquare$

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