Let $(X, M, \mu)$ be a measure space and $f: X\rightarrow \mathbb C$ be a essentially bounded measurable function (i.e. measurable and bounded on some $E\in M$ with $\mu(X\setminus E)=0$).

To define an essential range of $f$, let $V$ be the union of all open subsets $U$ of $\mathbb C$ such tat $\mu(f^{-1}(U))=0$ ( $V$ is the largest open subset of $\mathbb C$ such that $\mu(f^{-1}(U))=0$). By an essential range of $f$ we mean the set $R(f):=\mathbb C \setminus V$ (i.e. the smallest closed subset $R(f) \subset \mathbb C$ such that $f(x)\in R(f)$ for a.a. $x\in X$).

Let's define

$$ \|f\|_1=\max \{ |r|: r\in R(f)\}, \textrm{ (exist, because $R(f) $ is compact) }, $$ $$ \|f\|_2=\inf \{a\geq 0: \mu \{x\in X: |f(x)|>a\}=0 \} \textrm{ (finite, because $f$ is essentially bounded) }. $$

Since $\mu \{x\in X: |f(x)|>\|f\|_1 \}=0 $, we have $\|f\|_2 \leq \|f\|_1$.

Is it also $\|f\|_1 \leq \|f\|_2$ ?


For each $a > \Vert f \Vert_2$, we have $\mu(f^{-1}(\Bbb{C} \setminus \overline{B_a (0)}))= \mu(\{x \mid |f(x)| > a\}) = 0$.

This means $\Bbb{C} \setminus \overline{B_a (0)} \subset V$ and hence $$R(f) = \Bbb{C} \setminus V \subset \Bbb{C} \setminus (\Bbb{C} \setminus \overline{B_a (0)}) = \overline{B_a (0)}.$$

We conclude

$$ \Vert f \Vert_1 \leq \max \{|r| \mid r \in \overline{B_a (0)} \} = a. $$

As $a > \Vert f \Vert_2$ was arbitrary, we arrive at $\Vert f \Vert_1 \leq \Vert f \Vert_2$.


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