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In René Schilling's book Measures, Integrals and Martingales (second edition), a footnote on p. 116 says,

Problem 13.21 shows that $\mathcal L^\infty$ is the limit of $\mathcal L^p$ as $p\to\infty$.

This seems wrong to me. In particular, consider the unit function $f(x)=1$. Clearly $f\in\mathcal L^\infty$, but $f\notin\mathcal L^p$ for all $p\in[1,\infty)$. Perhaps Schilling means a different kind of set-theoretic limit from what I am expecting? My understanding is that we can define $\limsup$ and $\liminf$ for set sequences as

$$\liminf_{n\to\infty}=\bigcup_{n\geqslant 1}\bigcap_{j\geqslant n}A_j$$

$$\limsup_{n\to\infty}=\bigcap_{n\geqslant 1}\bigcup_{j\geqslant n}A_j$$

and then define the limit to be equal to these sets if they are equal to each other.

With this definition, $f(x)=1$ is in neither the $\limsup$ nor the $\liminf$, so Schilling's claim doesn't seem to hold up.

Contrary to the footnote, Problem 13.21 says nothing about this claim.

Note: This is a similar topic, but a distinct question, from that of this post, "Limit of $L^p$ Norm." That post is about the numerical limit of $\|\cdot\|_p$, whereas this question is about the set-theoretic limit of $\mathcal L^p$. And $\lim_{p\to\infty}\|\cdot\|_p=\|\cdot\|_\infty$ does not necessarily imply $\lim_{p\to\infty}\mathcal L^p=\mathcal L^\infty$.

WillG
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    See Limit of $L^p$ norm. – Dave L. Renfro Mar 16 '20 at 20:47
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    That post is about the numerical limit of $|\cdot|p$, whereas this question is about the set-theoretic limit of $\mathcal L^p$. And $\lim{p\to\infty}|\cdot|p=|\cdot|\infty$ does not imply $\lim_{p\to\infty}\mathcal L^p=\mathcal L^\infty$. – WillG Mar 16 '20 at 20:54
  • Sorry, I didn't look at it carefully enough to notice. I did this quickly as I was leaving for some errands away from home, which is also why I only posted one such link (usually when I make comments like that, I'll give several from Mathematics, Mathematics Educators, mathoverflow, etc.). When I saw your comment now I thought I'd probably delete my comment, but maybe it's better to leave my comment here along with your clarification. Perhaps even better, you could incorporate this clarification (and give the link to the other Stack Exchange question) near the beginning of your question. – Dave L. Renfro Mar 17 '20 at 07:45
  • @DaveL.Renfro Thanks, I've added this clarification and link to the question. – WillG Mar 17 '20 at 16:28
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    Problem $13.21$ is about the limit of the norms, so I assume that's what was meant. The statement in terms of set-theoretic limits is wrong and I don't think it can be salvaged. Even for a finite measure space, $\lim_{p\rightarrow\infty}\mathcal{L}^p$ exists (in the set-theoretic sense), but can be a proper superset of $\mathcal{L}^{\infty}$. Also, it appears you got the $\limsup$ and $\liminf$ definitions switched. – Thorgott Mar 17 '20 at 16:49
  • Actually, $\lim_{p\rightarrow\infty}\mathcal{L}^p$ always exists, but it neither need to contain nor be contained in $\mathcal{L}^{\infty}$. I'm not sure how much we can generally say about that limit. – Thorgott Mar 17 '20 at 16:59
  • @Thorgott Thanks, I corrected the lim sup/inf definitions. – WillG Mar 17 '20 at 17:04
  • $\mathcal{L}^{\infty}\subseteq\lim_{p\rightarrow\infty}\mathcal{L}^p\Leftrightarrow\mu(X)<\infty$. The reverse inclusion has something to do with the $p$-norms of functions not blowing up, but I don't know a good characterization thereof. – Thorgott Mar 17 '20 at 19:37

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