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It seems like different authors use different definitions for the space $L^\infty$. For example wikipedia starts with bounded functions, and define the seminorm $\lVert f \rVert_\infty$, and take quotients. When defining the seminorm, wikipedia takes the infimum over all $M \geq 0$ such that the set $\left\{x\in X\ |\ f(x)>M\right\}$ is null.

The books by Folland (Real Analysis) or Jones (Lebesgue Integration on Euclidean Space), on the other hand, starts with essentially bounded functions; $f$ is essentially bounded iff there is some $M\geq 0$ that makes $\left\{x\in X\ |\ f(x)>M\right\}$ a null set. Then they define the seminorm, take quotients, just like wikipedia.

The book by Cohn (Measure Theory) starts with bounded functions, but the seminorm differs! Here, the seminorm is given by the infimum over all $M\geq 0$ such that the set $\left\{x\in X\ |\ f(x)>M\right\}$ is locally null. When the given measure is $\sigma$-finite, the concept of locally null and null coincide, so this definition agrees with wikipedia's.

So the definition differs from books to books. There are at least 2*2=4 possibilities I guess;

  • Which seminorm do you use? ($\left\{x\in X\ |\ f(x)>M\right\}$ is null? or locally null?...)

  • What functions do you start with? (bounded functions? or those with finite seminorms?...)

Do the results concerning the properties of $L^p$ differs significantly when choosing a different definition for $L^\infty$ ? Do these differences affect the theory in a serious manner? Or, is it the case that, at least for familiar spaces such as $\mathbb{R}^n$ they all agree?

Thank you in advance.

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I have been asking myself the same question for some time, so let me share some ideas, which are far from constituting a full answer.

The definitions of Ash, Rudin and some others is $$\|f\|_\infty = \text{ess sup} |f| = \inf\{c: \mu\{|f|>c\} =0\}$$

while on the other hand, Cohn says $$\|f\|_\infty = \inf\{c: \{|f|>c\} \text{ is locally null} \}$$

The idea is to consider $L_\infty$ as the space of (equivalence classes of) functions for which $\|f\|_\infty<\infty$, regardless on which definition you use. With the norm $\|f\|_\infty$ this is a Banach space in any case.

Now, since a set $A$ is by definition locally null if $\mu(A\cap B)=0$ for every $B$ of finite measure, we can easily infer the following properties:

1) Every null set is locally null.

2) If $A$ is locally null and $B$ is $\sigma$-finite, then $A\cap B$ is null.

This gives us that

  • If $f\in L_\infty$ in the sense of Ash, then $f\in L_\infty$ in the sense of Cohn.
  • For $\sigma$-finite measure spaces, both definitions agree.

I don't know anything about more general spaces. There are some examples on Cohn (p. 98) of locally null but non null sets, but they take place in rather trivial or rather involved spaces.

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If the space $(X,\mathscr{B},\mu)$ is $\sigma$--finite the notion of locally null and null sets as well as the essential suprema and local essential suprema coincide:

For a given numerical measurable function $f$ with $\mu(|f|=\infty)=0$, define \begin{aligned} B_e&:=\{M>0: \mu(|f|>M)=0\}\\ B_\ell&:=\{M>0:\mu(A\cap\{|f|>M\})=0,\,\forall A\in\mathcal{L}_1\} \end{aligned}

Since $B_e\subset B_\ell$, $$\|f\|^\ell_\infty=\inf B_l\leq \inf B_e=:\|f\|^e_\infty$$ Hence, the essential supremum of $f$ is at least as large as the local essential supremum. It is easily checked that $\mu(|f|>\|f\|^e_\infty)= 0$, and $\mu(A\cap |f|>\|f\|^\ell_\infty)=0$ for any integrable set $A$.

In the case where $X=\bigcup_{\in\mathbb{N}}A_n$, $A_n\in\mathcal{L}_1$, $$ 0=\sum_n\mu(A_n\cap\{|f|>\|f\|^\ell_\infty)=\mu(|f|>\|f\|^\ell_\infty) $$ and so, $\|f\|^e_\infty\leq \|f\|^l_\infty\leq \|f\|^e_l$. This shows that for $\sigma$--finite spaces there is no need to worry about locally null sets for they are the same as null sets and nothing is gained by introducing the former.


One of the main reasons one deals with locally null sets when $\mu$ is not $\sigma$-finite is to guarantee that a version of Riesz Representation concerning the dual space of $(L_1)^*$, namely the linear map $T:L_\infty\rightarrow L^*_1$ is an isometry. A couple of things may go wrong if $(X,\mathscr{B},\mu)$ is not $\sigma$-finite space.

  1. Depending on which pseudonorm is used ($\|\;\|^\ell_\infty$ or $\|\;\|^e_\infty$) the definition of $L_\infty$ may exhibit quite different properties.

  2. Choosing $\|\;\|^\ell_\infty$ is better for duality. Even so, $T$ may failed to be onto.

Example 1: Consider a non empty set $X$ equipped with the trivial $\{\emptyset,X\}$ and the measure $\mu(X)=\infty$, $\mu(\emptyset)=0$. Then, the space $\mathcal{L}_0$ of all measurable functions coincides with the the space of all constant functions; a function $f$ is $\mu$-a.s 0 iff $f$ is the constant function $0$ and so, $\mathcal{L}_0=L_0$. For any $0<p<\infty$, $\mathcal{L}_p=\{\mathbf{0}\}$. The identification $\sim_p$ of functions $f,g$ such that $\|f-g\|_p=0$ gives $\mathcal{L}_p=L_p$ for all $0<p<\infty$. If $f\equiv c$, $\|f\|^\ell_\infty=0\leq |c|=\|f\|^e_\infty$. Hence $\mathcal{L}_\infty=\mathcal{L}_0$. The identification $\sim^\ell_{\infty}$ using $\|\;\|^\ell_\infty$ gives $L_\infty=\{\mathbf{0}\}$ while the corresponding $\sim^e_\infty$ using $\|\;\|^e_\infty$ gives $L_\infty=\mathcal{L}_\infty=\mathcal{L}_0$. This example shows, at least for duality purposes, why it is convenient to use $\|\;\|^\ell_\infty$.

Example 2: Cohn's measure theory textbook (p. 140, example 4.5.2) shows an example where $T$ is not surjective.


$L_p$ spaces ($1< p<\infty$) are not impacted, that because the Riesz representation ($L^*_p$ and $L_q$ are isometric, $\frac1p+\frac1q=1$) holds whether $\mu$ is $\sigma$-finite or not. For $p=1$ and $\mu$ non $\sigma$-finite, the isometry $L^*_1$ and $L_\infty$ holds when the space $(X,\mathscr{F},\mu)$ has additional structure (often topological).


Two other aspects of integration theory which are impacted outside the realm of duality for non-$\sigma$-finite spaces are (a) product of spaces (there may not be a Fubini-Tonelli type theorem) and (b) Radon--Nykodim derivative (there may not be one).

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