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I think $L^\infty(\Omega, \mathcal{F}, \mu)\supseteq L^p(\Omega, \mathcal{F}, \mu), \forall p \in (0, \infty)$? My reason is $L^\infty$ is defined as the set of measurable functions that are bounded up to a set of measure zero, and if $f \notin L^\infty$, then there exists a subset of measure nonzero on which $|f|$ is $\infty$, so $f \notin L^p, \forall p \in (0, \infty)$.

So I wonder why "If $\mu$ is finite, then $L^\infty(\Omega, \mathcal{F}, \mu)\subseteq L^p(\Omega, \mathcal{F}, \mu)$ for each $p$"? If we both are right, then If $\mu$ is finite, $L^\infty = L^p$?

This is a spinoff of a reply to my earlier question.

Thanks!

Community
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Tim
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    See Nate Eldredge's answer for the answer to the question in the title. A very simple question: can you give an example of an unbounded function in $L^1[0,1]$? – Martin Dec 28 '12 at 00:52
  • No. For $p=2$, the function $f(x) = x^{-1/4}$ is in $L^2[0, 1]$, but is not in $L^{\infty}[0, 1]$. – avs Apr 07 '19 at 19:49

2 Answers2

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The general result is described here: for any measure space $(S,\mu)$, and $0\leq p<q\leq \infty$,

  • $L^q(S,\mu)\subseteq L^p(S,\mu)$ $\iff$ $S$ does not contain sets of arbitrarily large measure
  • $L^p(S,\mu)\subseteq L^q(S,\mu)$ $\iff$ $S$ does not contain sets of arbitrarily small (non-zero) measure

Therefore, in particular, for a finite measure space $(S,\mu)$ we have $L^q(S,\mu)\subseteq L^p(S,\mu)$ for all $0\leq p<q\leq \infty$.

Zev Chonoles
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    Thanks! (1) Does "S does not contain sets of arbitrarily large measure" mean the same as "$\mu$ is finite"? (2) What does "S does not contain sets of arbitrarily small (non-zero) measure" imply about $\mu$? – Tim Dec 28 '12 at 03:19
  • (1): No, you can have a trivial example, like where $S$ consists of one point whose measure is infinite. Then $L^q$ and $L^p$ both consist of only the zero function. – Nate Eldredge Apr 19 '21 at 00:14
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If $f$ is not in $L_\infty$, it does not follow that $\{x:|f(x)|=\infty\}$ has positive measure. For example, the function $f:(0,1)\rightarrow\Bbb R$ defined by $f(x)=1/x$ is not in $L_\infty(0,1)$.

David Mitra
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