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Is there any rigorous or heuristic notion of boundary of $L^1$ that is studied? I mean something loosely like the collection of functions or distributions defined by

$$\left\{f\notin L^1: f_n\to f\quad\text{a.e.}\quad \text{as} \quad n\to \infty \quad \text{where} \quad f_n\in L^1\right\}$$

And what kind of characterizations or properties of this "surface" are known?

Edit: Changed to pointwise convergence.

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2 Answers 2

up vote 7 down vote accepted

I believe the set is just the set of all (non-integrable) measurable functions with $\sigma$-finite support.

In particular, if the space in question is $\sigma$-finite (like $\mathbf R$ with Lebesgue measure), then all measurable functions are pointwise limits of integrable functions.

Edit: cleaned up a bit.

In one direction:

  1. Clearly it is enough to show that for nonnegative functions;
  2. Let $f$ be a nonnegative measurable function, and $S_n$ an increasing sequence of sets of finite measure such that $\mathrm{supp}f\subseteq \bigcup_nS_n$.
  3. Let $A=\lbrace x\mid f(x)=\infty\rbrace$, $A_n=\lbrace x\mid f(x)<n\rbrace$.
  4. Then for each $n$ put $f_n(x)=f(x)$ on $A_n\cap S_n$, $f_n(x)=n$ on $A\cap S_n$, $f_n(x)=0$ otherwise.
  5. Then $f_n$ are integrable and $f_n\to f$ pointwise.

In the other direction, we show that the support of a pointwise limit of a sequence of integrable function has $\sigma$-finite support:

  1. Take an arbitrary sequence of integrable functions $f_n$
  2. Any integrable function $f_n$ has a $\sigma$-finite support $B_n=\bigcup_m\lbrace x\mid \lvert f_n(x)\rvert>1/m\rbrace$.
  3. The support of the limit of $f_n$ is contained in $\bigcup_n B_n$ (because if at some point none of the functions is nonzero, neither is the limit), and hence $\sigma$-finite, so we're done.
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Maybe it would be worth mentioning that you're answering the follow-up question in the comment to the other answer. –  t.b. Aug 6 '12 at 12:21
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@t.b.: No, I'm answering the question post-edit (which also happens to be the comment follow-up question...). The author has changed it. (I haven't even seen the initial version, I'm guessing it was about norm convergence.) Admittedly, completely erasing part of the content is not good practice (which is why I did not upvote the question, which is otherwise somewhat interesting...). –  tomasz Aug 6 '12 at 12:33
    
I see. I didn't notice the complete rewrite of the question (which was indeed about the norm)... But this makes me wonder why the other answer keeps getting upvotes while yours is essentially ignored. Oh well, voters here :) –  t.b. Aug 6 '12 at 12:39
    
@t.b.: I did the first upvote for the other answer because I thought that the answerer deserved credit for apparently answering the question as was first stated. As to why I'm not getting votes -- maybe others did not notice the edit, like you, or maybe they just don't bother to read the thread. I noticed that the highest scores I get for answers, I often get for the most trivial ones which are easy to understand for everyone. –  tomasz Aug 6 '12 at 13:02
    
I know what it means for a measure to be $\sigma-$finite. What does it mean for a measurable function to have $\sigma-$finite support? –  user782220 Aug 16 '12 at 6:00
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$L^1$ is Banach space, your set is therefore empty. You can also prove it in this way: for every $m,n$ we have $$ \int_\mathbb{R}|f_n-f_m| \le \int_\mathbb{R}|f_n-f|+\int_\mathbb{R}|f-f_m|, $$ i.e. $(f_n)$ is Cauchy sequence, and since $L^1$ is Banach space, there is some $g \in L^1$ such that $f_n \to g$. By uniqueness of the limit we conclude that $f=g \in L^1$.

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Ok what about using pointwise convergence instead? –  user782220 Aug 6 '12 at 1:54
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