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I am just taking a measure theory course and wondered:

Let $1 \leq p \leq q \leq \infty$. Does it always hold that

$$L^{\infty} \subseteq L^{q} \subseteq L^{p}?$$

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    Yes, if the underlying measure space is finite. – 2017-02-02
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    How would you go about proving that? – 2017-02-02
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    That is a little too long for a comment - I will give some hints in an answer. – 2017-02-02

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If $\mu(\Omega)<\infty$ yes, because $$\int_{\Omega}\lvert f\rvert^p\,d\mu\le\int_{\{\lvert f\rvert\le 1\}} 1\,d\mu+\int_{\{\lvert f\rvert>1\}}\lvert f\rvert^q\,d\mu\le \mu(\Omega)+\lVert f\rVert_q$$ In other cases, it might not. For instance, if $\mu$ is the counting measure on $\Bbb N$, then the inclusions are reversed. In $\Bbb R$ with Lebesgue measure, $(L^p\cap L^q)\setminus L^s$ is always non-trivial for any three distinct $p,q,s\in [1,\infty]$.

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    Thanks for your answer. Here you showed $L^{q} \subseteq L^{p}$, but what about $L^{\infty} \subseteq L^{q}$? – 2017-02-02
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    That's obvious. – 2017-02-02
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    It is not necessary that $\mu(\Omega) < \infty$ for the result to hold. – 2017-02-02
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    Why isn't it, Umberto P.? – 2017-02-02
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    Well, certainly for $L^\infty\subseteq L^p$ you need $\mu(\Omega)<\infty$. For $L^q\subseteq L^p$ perhaps no, but in many cases you won't have that inclusion. – 2017-02-02
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Suppose $(X,\cal M,\mu)$ is a finite measure space and $q < \infty$.

If $f \in L^\infty(\mu)$ then $$\int_X |f|^q \, d\mu \le \int_X \|f\|_\infty^q \, d\mu = \mu(X) \|f\|_\infty^q.$$

For the case $p < q$ see G. Sassatelli's answer.