Theorem. Let $p\in[1,\infty)$ and let $\{f_n\}_{n=1}^\infty\subseteq L^p(X,\mathcal{A},\mu)$ be a Cauchy sequence, then exists $f\in L^p$ such that $$\Large \left \| f_n-f\right \|_p\to 0 $$
Proof. By hypothesis we have $$\left(\forall\varepsilon > 0\right)\quad \left(\exists n(\varepsilon)\in\mathbb{N}\right)\quad\left(\forall m,n >n(\varepsilon)\right),\quad \left \|f_m-f_n\right \|<\varepsilon.$$ We know that exists a subsequence $\{f_{n_k}\}\subseteq\{f_n\}$ which converges almost everywhere in $X$. We define $$f:=\lim_{k\to\infty} f_{n_k}\quad\text{a.e.}.$$ Fixed $\varepsilon >0$ and $m>n(\varepsilon)$ results for the continuity of the absolute value and Fatou's Lemma that
\begin{equation}
\begin{aligned}
\int_X\left |f-f_m\right |^p\;d\mu=\int_X \left\{ \lim_{k\to\infty}\left | f_{n_k}-f_m\right|^p\right\}\;d\mu\le \liminf_{k\to\infty}\left \|f_{n_k}-f_m\right \|_p^p\color{red}{\le}\varepsilon^p
\end{aligned}
\end{equation}
Question. I don't understand the formal motivation of the inequality in red. Could anyone help me?
Best Answer
We're assuming "$m > n(\epsilon)$" in the line immediately before the calculations with the red inequality. What the author means by this is that we're applying the definition of cauchy-ness.
Remember that $\{ f_n \}$ is cauchy if and only if for every $\epsilon$, there is an $n(\epsilon)$ so that whenever $m, n > n(\epsilon)$, $\lVert f_n - f_m \rVert_p < \epsilon$.
Now in our calculation we're assuming that $m > n(\epsilon)$. But for large enough $k$, we know $n_k > n(\epsilon)$ too (since $n_k \to \infty)$.
That means for large enough $k$, we have both $m, n_k > n(\epsilon)$ so that cauchy-ness kicks in and we must have
$$\lVert f_{n_k} - f_m \rVert_p < \epsilon.$$
But raising both sides of this inequalty to the $p$ and taking the $\liminf$ as $k \to \infty$ (this is what guarantees that $k$ is "large enough") gives your red inequality.
I hope this helps ^_^