The space $L^1(\Omega)$ is never reflexive except in the trivial case where $\Omega$ consists of a finite number of atoms—and then $L^1(\Omega)$ is finite-dimensional.

this question is in functional analysis Brezis. why where $\Omega$ consists of a finite number of atoms then $L^1(\Omega)$ is finite-dimensional?


Let $A=\{x_1,\dotsc,x_n\}$ be our set with a finite number $n$ of atoms. Then if $f\in L^1(\Omega)$, it will have finite values on those atoms, i.e. $f(x_i)$ is finite for all $i$, since otherwise the integral would be infinite. But then $f$ is certainly $L^1$-equivalent to a function which is 0 for all $x\in\Omega\smallsetminus A$ and equal to it on those $n$ points. That is because the rest of $\Omega$ is a null set, and changing a function on a null set doesn't change its equivalence class in $L^1$ (or $L^p$ for any $p$ anyway). This means $L^1(\Omega)$ is generated by the indicators of the singletons of $A$, and is thus finite-dimensional.

Edit: As noted in comments below, if $f(x_i)=+\infty$ for some $i$, then the integral could not be finite, because: $$\int_\Omega f\mathrm{d}\mu=\sum_{i=1}^n\mu(\{x_i\})f(x_i),$$ so at least one term would be infinite. Same thing goes if $f(x_i)=-\infty$ for some $i$. Of course, having an infinite term doesn't imply the sum is infinite: it could be undefined. But we need it to be finite and exist, so that case is also unacceptable.

Edit 2: answer for dummies $\Omega$ is the domain of our functions, and our measure space. From the information you give, I have no reason to believe there is anything of positive measure outside the finite set of the atoms, which I called $A$ above. In fact, I have no reason to believe there is anything at all outside $A$ in $\Omega$, but I was trying to be more general :). If there is anything outside $A$ that has positive measure but is not an atom, then we can't continue.

Reading the article, I realized I thought of an atom in a wrong way. I thought of an atom as a singleton of positive measure, but that is not the case. The above answer more or less works anyway, but I will elaborate in a while.

So forget about the set $A$. $\Omega$ is a union of sets $A_i$ with positive measures which have no positive-measure subsets. Anything outside the $A_i$s is negligible. Those $A_i$ can be thought of as points, because the first step in studying a function here is to reduce it to a function constant on the $A_i$s. To do that, we first establish a little result.

Lemma If $A$ is a positive-measure atom of a measure $\mu$, then either there exists $B\subseteq A$ with $\mu(B)=\mu(A)$, or there are no non-empty proper measurable subsets of $A$.

Proof. Suppose we find an atom $A$ such that no proper subset of $A$ has the same measure of $A$. It follows that if $\varnothing\subsetneq B\subsetneq A$, then $\mu(B)=0$ or $B$ is not measurable. But then both $B$ and $A\smallsetminus B$ are measurable with measure 0, since they are both non-empty and proper subsets of $A$, and thus $A$ has measure 0 by finite additivity, since $\mu(A)=\mu(B)+\mu(A\smallsetminus B)$, which is absurd since we supposed $\mu(A)>0$. $\hspace{2cm}\square$

This means that if $A$ is a positive-measure atom, either we cannot measure anything smaller than it, or it has a subset with the same measure as $A$. So let us assume that our $A_i$ are sets with no measurable subsets. This means that a measurable function must be constant on them, otherwise there will be at least an $\alpha\in\mathbb{R}$ for which $f^{-1}(\alpha)$ is not measurable. Let me elaborate on this. If there exists an atom $A_i$ for which $f(A_i)$ is not a single element of $\mathbb{R}$, then $f(A_i)\supseteq\{x,y\}$ for some two $x,y\in\mathbb{R}$. But then $\varnothing\subsetneq f^{-1}(\{x\})\cap A_i\subsetneq A_i$, and being a nonempty proper subset of $A_i$ implies being nonmeasurable, so $f$ would not be measurable. This means that our function is necessarily a combination of indicators of $A_i$ on the atoms, and whatever it wants outside. In other words, it must be constant on the atoms.

Outside the atoms, everything has zero measure, which means what happens outside the atoms is of no interest to $L^1$. So assume $f(A_i)=\{x_i\}$, i.e. $f$ is constantly $x_i$ on $A_i$. Then: $$f\sim\sum_{i=1}^n\mathbb{1}_{A_i}\cdot x_i.$$ And this proves those indicators generate $L^1$. No $x_i$ can be infinite if $f\in L^1$, otherwise the integral would not be finite, because: $$\int_\Omega f\mathrm{d}\mu=\sum x_i\mu(A_i),$$ and if there is even a single nonfinite $x_i$, we have a nonfinite term in that sum, which can thus either be infinite, or not exist.

Hope this second edit finally makes everything clear.

Addendum A huge comment thread has been moved to chat from comments below. Besides the fact that this will pose a problem as my mobile just won't let me chat via Internet, someone might be curious as to what the comments said. So I made screenshots and here they are: 1, 2, 3, 4 and 5. Just in case.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Jun 17 '15 at 19:34
  • $\begingroup$ But the site won't let me chat. It says I must be logged in to chat... $\endgroup$
    – MickG
    Jun 17 '15 at 21:28
  • $\begingroup$ Oy. I thought you had finished the comment thread. But regardless, you are a registered user with way more reputation than necessary to chat. There should not be a problem. Isn't there a "Click here to log in" somewhere? [Sorry, haven't been logged out for so long, I forgot the details.] $\endgroup$ Jun 17 '15 at 21:36
  • $\begingroup$ Indeed I remember chatting with egreg once over at TeX SX. I'll check if it's the site or the mobile. What happens is that clicking on "logged in" in the above quoted phrase leads me to a page I will soon screenshot. The moved comment thread was so long I can't be sure whether it is finished or not :). $\endgroup$
    – MickG
    Jun 18 '15 at 5:16
  • $\begingroup$ Seems it's just my mobile. $\endgroup$
    – MickG
    Jun 18 '15 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.