Possible error in Donald Cohns measure theory on definition of $\mathscr{L}^{\infty}$ Let $(X,\mathscr{A},\mu)$ be a measure space. In the second edition of measure theory by Donald L. Cohn on page 92 the space $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ is defined to be the set of all bounded real-valued $\mathscr{A}$-measurable functions on $X$, not essentially bounded. More specifically he writes

Let $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ be the set of all bounded real-valued $\mathscr{A}$-measurable functions on $X$...

For a $\mu$-finite space $X$ his definition of the $\left\lVert \cdot\right\rVert_\infty$-norm is equivalent to
$$\left\lVert f\right\rVert_{\infty} = \inf\{M\mid \mu(\{x\in X\mid |f(x)|>M\}) =0\}.$$
In exercise 3.3.7 he then considers a finite measure space and a real-valued $\mathscr{A}$-measurable function and he proposes that $f\in \mathscr{L}^{\infty}(X,\mathscr{A},\mu)$ if and only if $f\in \mathscr{L}^{p}(X,\mathscr{A},\mu)$ and $\sup\{\left\lVert f\right\rVert_p\mid 1 \leq p <+\infty\}$ is finite.
I assert that the if statement is false. For if $X = [0,1]$ and we consider ordinary Lebesgue measure then if $\{q_n\}_{n=1}^{\infty}$ is an enumeration of the rationals in $[0,1]$ and
$$f = \sum_{n=1}^{\infty}n\chi_{\{q_n\}}$$
then $\left\lVert f\right\rVert_p = 0$ for all $p$ and $f\in \mathscr{L}^{p}$ but $f$ is certainly not bounded. 
Is this reasoning correct or am I missing something? I am assuming that for this question to be true he means that $f\in \mathscr{L}^{\infty}$ means that $f$ is essentially bounded for which I can prove that the two statements are equivalent. I am very thankful for any help.
 A: The set $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ is the set of all $\mathscr{A}$ - measurable functions $f:X \to \mathbb R$ , which are essentially bounded. Such a function need not to be bounded on all of $X$ !
Your function $f = \sum_{n=1}^{\infty}n\chi_{\{q_n\}}$ is not bounded on $[0,1]$, but we have
$f(x)=0$ for all $x \in [0,1] \setminus \{q_1,q_2,q_3,...\}$  and $\{q_1,q_2,q_3,...\}$ is a set of measure $0$, hence $f$ essentially bounded.
A: I am currently working through Cohn's Measure Theory, and I also thought that this was a bit odd. I certainly agree with the counter-example you gave. In the first edition of the book, he defines $\mathscr{L}^\infty(X,\mathscr{A},\mu,\mathbb{R})$ to be the set of all essentially bounded $\mathscr{A}$-measurable functions, whereas in the second edition he defines it as the set of all bounded $\mathscr{A}$-measurable functions. However, the exercise you mention is the same in both editions of the book. So, I guess that he just forgot to edit it when he changed the definition of $\mathscr{L}^\infty(X,\mathscr{A},\mu,\mathbb{R})$.
