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< Give an example of a finite measure $\mu$ on a \mb{} space~$X$
< and $f \in L^{\infty} (X, \mu)$ such that that $m (f)$ is injective.
< but not surjective.
---
> Give, with proof, an example of a finite measure $\mu$
> on a \mb{} space~$X$ and $f \in L^{\infty} (X, \mu)$
> such that that $m (f)$ is injective but not surjective.
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<  \leq \sum_{n = 1}^{\infty}
---
>  \leq \sup_{n \in \N}
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< in the proof of Proposition~\ref{4124_H6_8}, but it simplifies the proof.
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> in the proof of Proposition~\ref{4124_H6_8}, but it
> significantly simplifies the proof.
> We demonstrate with a direct proof of the implication from
> (\ref{4124_H6_8_1}) to~(\ref{4124_H6_8_4}).
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> \begin{proof}[Alternate proof
>  of (\ref{4124_H6_8_1}) imples~(\ref{4124_H6_8_4})
>  in Proposition~\ref{4124_H6_8}]
> Assume that~(\ref{4124_H6_8_4}) fails.
> Set $E = \bigl\{ x \in X \colon f (x) = 0 \bigr\}$.
> There are two cases.
> 
> First suppose that $\mu (E) > 0$.
> By semifiniteness, there is $F \S E$ with $0 < \mu (F) < \I$.
> Then $\ch_F \in L^p (X, \mu)$.
> However, $\ch_F \not\in {\operatorname{Ran}} (m (f))$,
> because for all $\xi \in L^p (X, \mu)$ we have $m (f) \xi |_F = 0$
> almost everywhere.
> 
> So suppose that $\mu (E) = 0$.
> For $n \in \N$ set
> \[
> S_n = \left\{ x \in X \colon
>   \frac{1}{n + 1} \leq | f (x) | < \frac{1}{n} \right\}.
> \]
> If there are only finitely many $n \in \N$ such that $\mu (S_n) > 0$,
> then there is $N \in \N$ such that $\mu (S_n) = 0$ for all $n \geq N$.
> Therefore
> \[
> \mu \left( \left\{ x \in X \colon
>   | f (x) | < \frac{1}{N} \right\} \right)
>  = \mu \left( \coprod_{n = N}^{\I} S_n \right)
>  = \sum_{n = N}^{\I} \mu (S_n)
>  = 0,
> \]
> contradicting the failure of~(\ref{4124_H6_8_4}).
> So there is a sequence $n (1) < n (2) < \cdots$ in $\N$
> such that for all $k \in \N$ we have $\mu (S_{n (k)}) > 0$.
> By semifiniteness, there is $T_k \S S_{n (k)}$ such that $0 < \mu (T_k) < \I$.
> 
> For $k \in \N$ set
> \[
> r_k = \frac{1}{k^{1 / (2 p)} n (k) \mu (T_k)^{1 / p} }.
> \]
> The sets $T_1, T_2, \ldots$ are disjoint.
> Set $T = \bigcup_{k = 1}^{\I} T_k$.
> Then we can define $\et \colon X \to [0, \I)$ by
> \[
> \et (x)
>  = \begin{cases}
>    r_k & \hspace*{1em} {\mbox{$k \in \N$ and $x \in T_k$}}
>        \\
>    0 & \hspace*{1em} x \in X \SM T.
> \end{cases}
> \]
> We have
> \[
> \int_X | \et |^p \, d \mu
>  = \sum_{k = 1}^{\I} r_k^p \mu (T_k)
>  = \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} n (k)^p}
>  \leq \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} \cdot k^p}
>  \leq \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} \cdot k}
>  < \I.
> \]
> so $\et \in L^p (X, \mu)$.
> 
> Suppose $\xi \in L^p (X, \mu)$ and $m (f) \xi = \et$.
> For $k \in \N$ and almost every $x \in T_k$, we then have
> \[
> | \xi (x) |
>   = \left| \frac{\et (x)}{f (x)} \right|
>   = \left| \frac{r_k}{f (x)} \right|
>   \geq r_k n (k).
> \]
> Therefore, using the Monotone Convergence Theorem,
> \[
> \int_X | \xi |^p \, d \mu
>  \geq \sum_{k = 1}^{\I} \int_{T_k} | \xi |^p \, d \mu
>  \geq \sum_{k = 1}^{\I} r_k^p n (k)^p \mu (T_k)
>  = \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2}}
>  = \I.
> \]
> This contradicts $\xi \in L^p (X, \mu)$.
> So $m (f)$ is not surjective.
> \end{proof}
> 
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< 
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> However, $m (f)$ is not surjective, by the criterion in
> Proposition \ref{4124_H6_8}(\ref{4124_H6_8_4}).
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< So the Uniform Boundedness Principle
---
> Since $E^*$ is complete, the Uniform Boundedness Principle
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< Therefore $\sup_{n \in \N} \| \xi_n \| < \infty$.
---
> So $\sup_{n \in \N} \| \xi_n \| < \infty$.