I encountered an exercise in my functional analysis course:
Let $(\Omega, \Sigma, \mu)$ be a measure space, and $A\subseteq L^p (\Omega, \mu) \cap L^r (\Omega ,\mu)$, where $1\leq p <r <+\infty$. If $A$ is a relatively compact subset of $L^p(\Omega,\mu)$, and $\forall \varepsilon>0$, $\exists M>0$, such that $$\int_{\left\{x\in \Omega : |f(x)| \geq M\right\}} |f(x)|^r \text{d} \mu (x) <\varepsilon, \qquad \forall f\in A. \label{eq4.1} \tag{$\star$}$$ Prove that $A$ is relatively compact in $L^r(\Omega, \mu)$.
Here is my answer:
Proof. We proceed in three steps:
Step 1. We claim that:
$A$ is $r^\text{th}$ absolutely equicontinuous in $L^r(\Omega, \mu)$. Here
$r^\text{th}$ absolutely equicontinuous means: $\forall \varepsilon>0$,
$\exists\delta>0$, such that
$$\left[\text{$E\subseteq \Omega$ and $\mu(E) <\delta$}\right]\Rightarrow
\left[\text{$\int_E |f|^r \text{d} \mu <\varepsilon \ $ for every $\ f\in A$}\right].\tag{1}\label{eq1}$$
Indeed, let $\varepsilon>0$ be arbitrary. By eq. \eqref{eq4.1}, we can choose $M>0$ such that
$$\int_{\{|f| \geq M\}} |f|^r\text{d}\mu <\frac{\varepsilon}{2}.$$
Taking $\delta = \frac{\varepsilon}{2M^r}$, then if $E\subseteq \Omega$ and $\mu(E) <\delta$, we have
\begin{align*}
\int_E |f|^r \text{d}\mu& =\int_{E\cap \{|f| <M\}} |f|^r \text{d}\mu +\int_{E\cap \{|f| \geq M\}} |f|^r \text{d}\mu\\
& \leq M^r \cdot \mu(E) + \frac{\varepsilon}2<\varepsilon.
\end{align*}
This proves eq. \eqref{eq1}.
Step 2. Let $\{f_n\}_{n=1}^\infty\subseteq A$ be an arbitrary sequence. Then there exists a subsequence $\{f_{n_k}\}_{k=1}^\infty$ converges to some $f$ in $L^p (\Omega,\mu)$. To simplify the notation, we write $\{f_k\}_{k=1}^\infty$ instead of $\{f_{n_k}\}_{k=1}^\infty$. By Chebyshev's inequality, we obtain that $\forall \varepsilon>0$, $$\mu(\{|f_k-f| \geq \varepsilon\}) \leq \frac{\|f_k-f\|_p^p}{\varepsilon^p}\to 0 , \qquad k\to +\infty,$$ $$\mu(\{|f_{k_1}-f_{k_2}| \geq \varepsilon\}) \leq \frac{\|f_{k_1}-f_{k_2}\|_p^p}{\varepsilon^p}\to 0 , \qquad k_1, k_2\to +\infty,$$ Thus $\{f_k\}_{k=1}^\infty$ converges in measure and is a Cauchy sequence in measure.
Step 3. Let $\varepsilon>0$ be arbitrary.
Since $\{f_k\}_{k=1}^\infty$ is a Cauchy sequence in $L^p(\Omega,\mu)$, we can find some integer $N_1>0$ such that
$$\|f_{k_1}-f_{k_2}\|_{p}^p<\frac{\varepsilon}3, \qquad \forall k_1,k_2 >N_1.$$
It follows from Step 1 that there exists some $\delta >0$ such that if $E\subseteq \Omega$ and
$\mu(E)<\delta$, we have
$$\left\| \chi_E\cdot f\right\|_r = \left(\int_E |f|^r \text{d}\mu\right)^{\frac1r}<\frac12\left(\frac{2\varepsilon}3\right)^{\frac1r}, \qquad \forall f\in A.$$
Then by Step 2, there exists some integer $N_2>0$, such that
$$\mu \left( \{ |f_{k_1} -f_{k_2}| \geq 1\}\right) <\delta, \qquad \forall k_1,k_2 >N_2.$$
Take $N=\max\{N_1,N_2\}$. Thus, for $k_1,k_2>N$,
\begin{align*}
\|f_{k_1}-f_{k_2}\|_r^r &% =\int_{\Omega \cap \left\{\left|f_{k_1} -f_{k_2}\right| <1\right\}} |f_{k_1}-f_{k_2} |^r \text{d}\mu +
= \left\|\chi_{\left\{\left|f_{k_1} -f_{k_2}\right| <1\right\}}\cdot (f_{k_1}-f_{k_2})\right\|_r^r+
\left\|\chi_{\left\{\left|f_{k_1} -f_{k_2}\right| \geq 1\right\}}\cdot (f_{k_1}-f_{k_2})\right\|_r^r\\
& \leq \|f_{k_1}-f_{k_2} \|_p^p+
\left( \left\|\chi_{\left\{\left|f_{k_1} -f_{k_2}\right| \geq 1\right\}}\cdot f_{k_1}\right\|_r+
\left\|\chi_{\left\{\left|f_{k_1} -f_{k_2}\right| \geq 1\right\}}\cdot f_{k_2}\right\|_r\right)^r\\
& <\frac{\varepsilon}3+\frac{2\varepsilon}3= \varepsilon.
\end{align*}
Therefore, $\{f_k\}_{k=1}^\infty$ is a Cauchy sequence in $L^r(\Omega,\mu)$, which completes the proof.$\qquad\qquad\blacksquare$
Is my answer correct? Moreover, I wonder whether there's different way to prove it, for example, by proving that $A$ is totally bounded (i.e., for any $\varepsilon>0$, there exist finitely many $\varepsilon$-balls covering $A$)?