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Suppose that $(X,S,\mu)$ is a measure space with $\mu(X)<\infty$. Suppose that $f:X\to [0,\infty)$ is $S-$ measurable. Suppose that $q>r$ are given positive numbers. Then, $\int f^rd\mu<\infty\implies \int f^q d\mu<\infty.$

I tried to prove it the following way: For brevity, let $\{f>k\}$ stand for the set $\{x\in X: f(x)>k\}$ etc.

Define $E_1:= \{f\lt 1\}, E_2=\{f=1\}, E_3=\{f>1\}$. $E_i\in S$ as $f$ is $S-$ measurable.

$\int_{E_1}f^qd\mu\le\int_{E_1} d\mu=\mu(E_1)\le \mu(X)<\infty$

$\int_{E_2}f^q=\int_{E_2}=\mu(E_2)\lt \infty$

$\int_{E_3}f^q\le\int_{E_3}f^r\le \int f^r<\infty$

It follows that $\int f^q=\int_{E_1}f+\int_{E_2}f+\int_{E_3}f<\infty$.

Is my proof correct? Thanks.

Koro
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    Your proof looks fine to me. But there is no need to introduce the set $E_2$, just consider $E_1 = { f\leq 1}$, then $\int_{E_1} f,\text{d}\mu \leq \mu(E_1) < \infty$. – stange May 29 '23 at 11:18
  • @stange: yeah, that works too. Thanks. – Koro May 29 '23 at 11:23
  • The proof works. As you know Holder's inequality will give you another answer with nice bounds. This may be of interest to you too – Mittens May 29 '23 at 16:15

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