2

I am studying Donald Cohn's Measure Theory. In Chapter 3, Exercise 7, the author asks to do the following exercise:


Let $(X, \mathcal A , \mu)$ be a finite measure space, and let $f$ be an $\mathcal A$ measurable real or complex valued function on $X$.

Show that $f$ belongs to $\mathscr L ^ \infty$ iff

  1. $f$ belongs to $\mathscr L^p (X,\mathscr A , \mu)$ for each $p \in [1, \infty )$ and
  2. $\sup \{ \lVert f \rVert _p : 1\le p < +\infty \}$ is finite.

Cohn defines $\mathscr L^p$ for $1\le p < \infty$ in the usual fashion. However, $\mathscr L ^\infty$ is the collection of all bounded measurable functions (this is different from what Wikipedia and other textbook do) and $\lVert f \rVert _\infty$ is defined to be the infimum of those nonnegative numbers $M$ such that $\{ x\in X : |f(x)| > M \}$ is locally null. (See here the definition of locally null)

I successfully proved the "only if" part. To prove the "if" part, I need to prove that if any measurable function on a finite measure space which satisfies conditions $1$ and $2$ of the question then it must be bounded.

However, I have a counterexample. Let's consider $X= (0,1]$, $\mathscr A$ is the Borel sigma algebra on $X$ and $\lambda$ is the Lebesgue measure on $X$. Consider the function $f$ on $X$ given by

$$f(x)= \begin{cases} n & \text{if } x=m/n \text{ with } \gcd(m,n)=1 \newline 0 & \text {otherwise} \end{cases}$$

Notice that $f$ is zero almost everywhere and $f$ is measurable because $f= \sum_{p/q \in \mathbb Q \cap (0,1]} q\chi_{\{ p/q\}}$ (and hence is a limit of simple measurable functions). But this function satisfies both conditions 1 and 2 however is not bounded.


Is my counterexample correct? If it is, can the hypothesis of the question be tweaked so the the assertion becomes correct?

meth-
  • 265
  • 1
    I pencilled a note in the margin of my copy (Exercise 3.3.7, page 98) last year: "This is only correct if $\mathscr{L}^\infty(X, \mathscr{A}, \mu)$ is defined as in the footnote on page 92, not as in the text itself." – Calum Gilhooley Nov 03 '22 at 15:43
  • The Maths.SE website has automatically found a near-duplicate question here: Possible error in Donald Cohns measure theory on definition of $\mathscr{L}^{\infty}$. – Calum Gilhooley Nov 03 '22 at 15:54
  • In the case of a finite or $\sigma$-finite measure $\mu$, $|f|\infty=\inf{M>0: {|f|>M},\text{is locally null}}=\inf{a>0:\mu(|f|>a)=0}$. Try to prove that yourself. In this posting there is a proof. Thus $f\in L\infty(\mu)$, $\int|f|^p,d\mu\leq|f|^p_\infty,\mu(X)<\infty$. – Mittens Nov 03 '22 at 15:55
  • Indeed your function is not bounded, but it is essentially bounded. $L_\infty$ is the soace of functions which outside sets of measure zero, are bounded. – Mittens Nov 03 '22 at 16:51
  • @CalumGilhooley thank you for the response. I sent a mail to the author regarding this. Hopefully, things will be fixed in the next edition of the textbook. – meth- Nov 04 '22 at 12:25
  • @meth-: there is not error in the particular statement you are quoting. – Mittens Nov 04 '22 at 14:28
  • 1
    @OliverDíaz I think the author should have specified which definition to use in the exercise as the definition in the text and the footnote are not equivalent. – meth- Nov 04 '22 at 17:41
  • @meth-:Probably he meant for the reader to see that in the setting of finite measure, both definitions of $L_\infty$ coincide. his footnote is very clear. he is using the one based on local null sets (specially for his presentation of liftings). – Mittens Nov 04 '22 at 17:50
  • 1
    @OliverDíaz the example that I give here clearly shows that the two definitions are not equivalent. Essentially bounded does not imply bounded even in a finite measure space. – meth- Nov 05 '22 at 07:50
  • Your example is clearly wrong. Your function there is indeed not bounded, but it is essentially bounded: $\lambda(|f|>0)=\lambda(\mathbb{Q})=0$. – Mittens Nov 05 '22 at 10:54
  • Check also this posting – Mittens Nov 05 '22 at 12:08
  • Why the downvote? Quoting from Why is voting important?: "Voting down a post signals $\ldots$ that the post contains wrong information, is poorly researched, or fails to communicate information." In what way have any of those criteria been met in this case? – Calum Gilhooley Dec 02 '22 at 06:56

1 Answers1

2

Let us check whether we are all on the same page (literally!).

From page 92 of Donald L. Cohn, Measure Theory (second edition 2013):

Let $\mathscr{L}^\infty(X, \mathscr{A}, \mu, \mathbb{R})$ be the set of all bounded real-valued $\mathscr{A}$-measurable functions on $X,$ and let $\mathscr{L}^\infty(X, \mathscr{A}, \mu, \mathbb{C})$ be the set of all bounded complex-valued $\mathscr{A}$-measurable functions on $X.$ [$\ldots$] In discussions that are valid for both real- and complex-valued functions we will often use $\mathscr{L}^p(X, \mathscr{A}, \mu)$ to represent either $\mathscr{L}^p(X, \mathscr{A}, \mu, \mathbb{R})$ or $\mathscr{L}^p(X, \mathscr{A}, \mu, \mathbb{C}).$

Footnote on same page:

Some authors define $\mathscr{L}^\infty(X, \mathscr{A}, \mu, \mathbb{R})$ and $\mathscr{L}^\infty(X, \mathscr{A}, \mu, \mathbb{C})$ to consist of functions $f$ that are essentially bounded, which means that there is a nonnegative number $M$ such that $\{x \in X : |f(x)| > M\}$ is locally $\mu$-null [$\ldots$]. For most purposes, it does not matter which definition of $\mathscr{L}^\infty$ one uses. [$\ldots$]

Main text, continued from page 92 to page 93:

We can define $\|\cdot\|_p$ in the case where $p = +\infty$ by letting $\|f\|_\infty$ be the infimum of those nonnegative numbers $M$ such that $\{x \in X : |f(x)| > M\}$ is locally $\mu$-null. Note that if $f \in \mathscr{L}^p(X, \mathscr{A}, \mu),$ then $\{x \in X : |f(x)| > \|f\|_\infty\}$ is locally $\mu$-null, for if $\{M_n\}$ is a nonincreasing sequence of real numbers such that $\|f\|_\infty = \lim_nM_n$ and such that for each $n$ the set $\{x \in X : |f(x)| > M_n\}$ is locally $\mu$-null, then the set $\{x \in X : |f(x)| > \|f\|_\infty\}$ is the union of the sets $\{x \in X : |f(x)| > M_n\}$ and so is locally $\mu$-null. Thus $\|f\|_\infty$ is not only the infimum of the set of numbers $M$ such that $\{x \in X : |f(x)| > M\}$ is locally $\mu$-null but is itself one of those numbers.

Exercise 3.3.7, on page 98:

Let $(X, \mathscr{A}, \mu)$ be a finite measure space, and let $f$ be an $\mathscr{A}$-measurable real- or complex-valued function on $X.$

  • (a) Show that $f$ belongs to $\mathscr{L}^\infty(X, \mathscr{A}, \mu)$ if and only if:

    • (i) $f$ belongs to $\mathscr{L}^p(X, \mathscr{A}, \mu)$ for each $p \in [1, +\infty),$ and

    • (ii) $\sup\{\|f\|_p : 1 \leq p < +\infty\}$ is finite.

  • (b) Show that if these conditions hold, then $\|f\|_\infty = \lim_{p \to +\infty}\|f\|_p.$

[I don't know how to nest lists properly in a blockquote in Markdown. Feel free to correct my formatting.]

There would be nothing wrong with the exercise if $\mathscr{L}^\infty(X, \mathscr{A}, \mu)$ were defined as in the footnote, but it is defined differently in the main text, and according to that definition, the OP's counterexample is valid.

  • It does seem that in the new addition of the book they committed the adjective essentially bounded before bounded in the definion of $\mathcal{L}\infty(X)$ in the paragraphs you wrote in your posting. However, keep reading past that paragraph and the author makes clear how the $\mathcal{L}\infty(X)$ and its nor mare defined (without need to the footnote). Of course, the author means, as in any other measure theory textbook, to consider properties almost surely. The OP was failed to understand that, but I hope by now it is obvious. – Mittens Nov 07 '22 at 23:52