There is a result in the wikipedia article about $L_p$ space embedding:

a. Let $0 ≤ p < q ≤ ∞$. $L_q(S, μ)$ is contained in $L_p(S, μ)$ iff $S$ does not contain sets of arbitrarily large measure;
b. Let $0 ≤ p < q ≤ ∞$. $L_p(S, μ)$ is contained in $L_q(S, μ)$ iff $S$ does not contain sets of arbitrarily small non-zero measure.

I can only find one which is similar to the result above: Another note on the inclusion $L^p(\mu) ⊂ L^q(\mu)$(by A. Villani, The American Mathematical Monthly, Vol. 92 (1985), No. 7, 485–487):

The following conditions on measure space $(X,\Sigma,\mu)$ are equivalent:
1. $\sup_{E\in{\mathscr A}_{\infty}}\mu(E)<+\infty$,where ${\mathscr A}_\infty=\{E\in\Sigma:\mu(E)<+\infty\}$.
2. $L^p(\mu)\subset L^q(\mu)$ for all $p,q\in(0,\infty)$ with $p>q$.

and

the following conditions on measure space $(X,\Sigma,\mu)$ are equivalent:
3. $\inf_{E\in{\mathscr A}_{0}}\mu(E)>0$,where ${\mathscr A}_0=\{E\in\Sigma:\mu(E)>0\}$.
4. $L^p(\mu)\subset L^q(\mu)$ for all $p,q\in(0,\infty]$ with $p<q$.

Is there a typo in (a) ($0\leq p<q\leq\infty$ should be $0\leq p<q<\infty$?)? Can someone come up with a reference for the results (a) and (b)?

  • no, there's no typo. Villani doesn't even define what a $L^0$ space is. (As an aside, what does $L^0$ mean for a general measure space to you?) – Willie Wong Mar 26 '13 at 16:56
  • @Willie, I've never seen $L^0$ in any textbook before. It is defined in the Wikipedia article en.wikipedia.org/wiki/…. – Jack Mar 26 '13 at 19:02

(This would be a comment if I could make one.)

There is a similar statement in Exercise 5 page 186 of Folland's Real Analysis. The statements there are the following:

Suppose $0 < p < q < \infty$. Then $L^p \not\subseteq L^q$ iff $X$ contains sets of arbitrarily small positive measure, and $L^q \not\subseteq L^p$ iff $X$ contains sets of arbitrarily large finite measure.

He gives some hints for the 'if' directions. (These statements are of course equivalent to the non-endpoint versions of your statements.)

On a different note, my guess for the definition of $L^0$ is $$ L^0 = \{\text{measurable } f : \mu(\text{supp} f) < \infty\} $$ where $\text{supp} f = \{x : f(x) \neq 0\}$ is defined up to a set of measure zero (and hence $\mu(\text{supp} f)$ is unambiguous).

(Edited in view of the comment of grew). I guess the following is true, which helps proving the statements:

$S$ does not contain sets of arbitrarily large measure: Membership in $L^p$ is decided as in $L^p(S^1)$, i.e., by the large values: $$f\in L^p \iff \max\lbrace |f|, 1\rbrace \in L^p.$$

$S$ does not contain sets of arbitrarily small non-zero measure: Membership in $L^p$ is decided as in $\ell^p$, i.e., by the small values (but this holds only if the large values do not disturb, e.g., for sequences in $c_0$).

Edit: This first paragraph is wrong: I think (a) is fine, because $1 \in L^p$ for $p \in [0,\infty]$ and therefore $f \in L^\infty$ implies $f \in L^p$.

But I think, there is an issue with (b) as it is contained in wikipedia, if $L^0$ are just the measureable functions (as wiki says): Take $S = \mathbb{N}$ with the counting measure. Then all functions (sequences) are measureable (hence in $L^0$), but may not be in any $L^p$, $p > 0$.

Together with Villani's result, there should be no open case left?

Sorry for the following comment, but I can't add comments with my reputation: Note that the second part of Peter Michor's answer is not correct, $f\in L^p$ always implies $\min(|f|,1) \in L^p$ without assumptions on $S$.

Edit: We have slipped the following pathological case: Let $X ={0}$ and $\mu(X) = \infty$. Then, $1 \not\in L^\infty$. This is the reason that Villani has excluded $p = +\infty$ in the second statement.

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