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I'm trying to prove that if $(X,\mu)$ is a measure space then $L^q(X,\mu)\subset L^p(X,\mu)\iff X$ does not contain sets of arbitrarily large measure, for $1\leq p<q\leq \infty$. I've managed to prove that if $X$ does contain such sets, then the inclusion is false, but I'm having trouble proving the converse, that is, if there is $f\in L^q\backslash L^p$ then $X$ contains sets of arbitrarily large measure. I've tried to use a similiar argument as the one for proving that $L^p\not\subset L^q\implies X$ contains sets of arbitrarily small positive measure, by analyzing the sets $\{x\in X||f(x)|>n\}$, or in this case their complementary, but I can't guarantee their measure isn't infinite. Any suggestions?

MathNewbie
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1 Answers1

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If $X$ does not contain sets of arbitrarily large measure then $\mu (X) <\infty$. [There exits $M<\infty$ such that $\mu (E) \leq M$ for all $E$]. But the Holder's inequality shows that $(\int |f|^{p}d\mu)^{1/p}\leq (\int |f|^{q}d\mu)^{p/q}{(\mu (X)}^{1-\frac p q}$ which finishes the proof.

geetha290krm
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  • Not necessarily, if $X={0,1}$ with $\mu({0})=\infty$ and $\mu({1})=0$, then $X$ does not contain sets of arbitrarily large measure. (From what I understand, we require them to be finite) – MathNewbie Jul 15 '22 at 23:49
  • In your example for every $n$ there exists a set $A$ with $\mu (A) >n$ (namley $A={0}$). I think you are reading 'contains sets of arbitrarily large measure' as 'contains sets of arbitrarily large finite measure'. @MathNewbie – geetha290krm Jul 15 '22 at 23:52
  • yes, I believe that is the correct interpretation, for if you consider my example, then $L^\infty = L^p = {f : f(0) = 0}$ so the inclusion holds. – MathNewbie Jul 18 '22 at 01:12