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Let $(X,\mathfrak{M},\mu)$ be a measure space. Show that if $\mu(X) < \infty$ then $L^q(X) \subset L^p(X)$ for $1\le p \le q \le \infty$.

Is it enough to define $$E_0 = \{x \in X \, : \, 0 \leq |f(x)| < 1\}$$ and $E_1 = X \setminus E_0$ so that $\mu(E_0), \, \mu(E_1) < \infty$ and $E_0$ is measurable as $E_0 = \{x \in X : |f(x)| \geq 0 \} \cap \{x \in X : |f(x)| \leq 1 \}$ and $f$ is measurable. Therefore if $f \in L^q$,

\begin{eqnarray*} ||f||^p_{L^p} &=& \int_{E_0}|f|^p \, d\mu + \int_{E_1} |f|^p \, d\mu \\ &\leq& \int_{E_0}|f|^p \, d\mu + \int_{E_1} |f|^q \, d\mu \\ &<& \mu(E_0) \,\, + ||f||_{L^q}^q \\ &<& \infty. \end{eqnarray*}

which implies $f \in L^p$.

Doe this suffice? I've seen a proof using Holder's inequality (that has now been given as an answer for those interested), that is much shorter, but does this suffice?

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    $\begingroup$ This looks good. $\endgroup$ – Sinister Cutlass Oct 18 '15 at 20:44
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    $\begingroup$ $E_0$ is measurable, and your work is good. But the standard way to prove is to apply Hölder's inequality. $\endgroup$ – Julián Aguirre Oct 18 '15 at 20:56
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    $\begingroup$ @JuliánAguirre Thank you. This was homework, and I didn't quite see the application of Holder. What guarantees $E_0$ is measurable? $\endgroup$ – Anthony Peter Oct 18 '15 at 20:58
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    $\begingroup$ @JuliánAguirre My apologies, of course it's measurable.. $f$ is measurable and $E_0$ is the intersection of the two level sets of $f$ $\endgroup$ – Anthony Peter Oct 18 '15 at 21:03
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    $\begingroup$ my silly mistake. thanks guys $\endgroup$ – R.N Oct 18 '15 at 21:10
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Your proof is fine. You can also try to show that if $p<q$ then $$\lVert f\rVert_p \leqslant \lVert f\rVert_q \mu(X)^{1/p-1/q}$$

by using Hölder's inequality. This gives a bit more information than your proof. In particular for $X$ a probability space gives $p\mapsto \lVert f\rVert_p$ is increasing.

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