$f$ is bounded and in $L^p(\Omega,\mathcal{A},\mu)$. Then $f\in L^q(\Omega,\mathcal{A},\mu), 1\leq p<q.$

Question

$(\Omega,\mathcal{A},\mu)$ is a measure space. $f:\Omega\rightarrow\mathbb{R}$ is bounded and in $L^p(\Omega,\mathcal{A},\mu)$.($\mu(\Omega)<\infty$ or Not) Then $f\in L^q(\Omega,\mathcal{A},\mu)$ for $1\leq p<q.$

$\int_\Omega|f|^q=\int_{\{|f|\leq1\}}|f|^q+\int_{\{|f|>1\}}|f|^q\leq\int_{\{|f|\leq1\}}|f|^p+\int_{\{|f|>1\}}|f|^q$. Since $f\in L^p$, the first term is finite. Can we also show the second is finite using the boundedness of $f$?

Answer 1

Since $|f|^{q}=|f|^{p}|f|^{q-p}$ we get $\int_{|f|>1} |f|^{q} \leq M^{q-p} \int_{|f|>1} |f|^{p} \leq M^{q-p} \int |f|^{p}$ where $M=\sup |f|$.

Answer 2

The second term is indeed finite because of boundedness of $f$. Note that $\mu(\{|f| > 1\})$ is finite since $\{|f| > 1\} = \{|f|^p > 1\}$ and $|f|^p$ is integrable (here we don't need that $f$ is bounded). Also since $|f|$ is bounded then $|f|^q$ is also bounded and therefore it is integrable on a set $\{|f| > 1\}$ because it has finite measure.