What does "sets of arbitrarily large measure" mean --- question about $L_p$ embeddings The following are the results from a wikipedia article about $L_p$ space: 

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.

Here are my questions:


*

*What does "sets of arbitrarily large measure" mean? Is "$S$ does not contain sets of arbitrarily large measure" equivalent to "$\mu(S)<+\infty$"?

*What does "sets of arbitrarily small non-zero measure" mean? 



[Added:]
There is a result in 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 are equivalent: 



*

*$\sup_{E\in{\mathscr A}_{\infty}}\mu(E)<+\infty$

*$L^p(\mu)\subset L^q(\mu)$ for all $p,q\in(0,\infty)$ with $p>q$


where ${\mathscr A}_\infty=\{E\in\Sigma:\mu(E)<+\infty\}$.
This is similar with (a) but $p,q\in(0,+\infty)$.
 A: Pick a positive number $r$. Does $S$ have sets with measure bigger than $r$? Deos $S$ have sets of measure smaller than $r$ but not zero?
If, for every value of $r$ you answered yes to the first question, $S$ contains sets of arbitrarily large measure.
If, for every value of $r$ you answered yes to the second question, $S$ contains sets of arbitrarily small, nonzero measure.
A: This is more of a comment response to Jack, but I need the formatting given by MarkDown.

The problem is that the Wikipedia sentence is ambiguous: and in one way of reading it, the statement is incorrect. Wikipedia contains:

Let $0\leq p\leq q\leq \infty$. $L^q(S,\mu)$ is contained in $L^p(S,\mu)$ iff $S$ does not contain sets of arbitrarily large measure.

A big spot where it is entirely unclear is whether the phrase "Let $0\leq \ldots$" is part of the "left-hand-side" of the "iff". The usage of a full stop . instead of a comma , would suggest that it is not.
Then if we interpret "does not contain..." as what you wrote in your question, that is $\mu(S) < \infty$, then the statement is false! Because as in my comment there exists $p,q\in (0,\infty)$ and a measure space $(S,\mu)$ such that $\mu(S) = +\infty$ while the $L^p \subset L^q$ inclusion holds, contradicting the "only if" of the "iff".
If we interpret "does not contain..." as what Villani wrote, that is $\sup_{A_\infty} \mu(E) < +\infty$, it is clear that we will have a problem with the endpoints. So neither interpretation works!

The only way to fix the statement is to correctly place your quantifiers. That is to say, you can write that

Given a measure space $(S,\mu)$, the following two conditions are equivalent:

*

*$\mu(S) < \infty$

*For every $0 < q \leq p \leq \infty$ the inclusion $L^p \subseteq L^q$ holds.


I do not include the endpoint $0$ as I do not know how you want to define the $L^0$ space when $\mu(S) = + \infty$. To show that $1\implies 2$ you just use Hölder/Jensen; to show that $2\implies 1$ just take the function $f\equiv 1$ which is in $L^\infty$. Integrating in any $L^p$ for $p <\infty$ tells you that $S$ has finite measure.
