In $(X,S,\mu), \int f^r d\mu<\infty\implies \int f^q d\mu<\infty$, where $\mu(X)<\infty$.

Question

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.