For $1\le p

I have some doubts about the proof of this theorem. From time to time I will put my justification.

For $1\le p < +\infty$, $L^p$ is a Banach space

Let $(f_n)_n$ be a Cauchy sequence in $L^p$, i.e., $(f_n)$ has the property that $\lim_{m,n\to \infty} \lVert f_n-f_m\rVert_p=0.$ We can find a subsequence of $(f_n)$ that does converge $\mu-$almost averywhere. In fact, chosse $(f_{n_k})_k$ as any subsequence of $(f_n)$ such that $n_1<n_2<\cdots<n_n<\cdots$ and $\sum_{k=1}^\infty \lVert f_{n_{k+1}}-f_{n_k}\rVert_p=\alpha<\infty$ This is possiblr: e.g., we can select increasing $n_k's$ such that $\lVert f_m-f_{n_k}\rVert_p<2^{-k}$ for all $m\ge n_k$.

My justification for this fact: For the definition of Cauchy sequence we have that $$(\forall k\in\mathbb{N})\quad (\exists n(k)\in\mathbb{N})\quad (\forall m,n \ge n(k))\quad \lVert f_m-f_n \rVert_p<2^{-k}$$I choose $n_1\ge n(1)$, then from the above definition results $\lVert f_m - f_{n_k}\rVert_p<2^{-1}$ for all $m\ge n_1$.

Now, I choose $n_2:=\max\{n_1+1, n(2)\}$, then $n_2>n_1$ and in particular $n_2\ge n(2)$ then from the definition we have that $\lVert f_m-f_{n_2}\rVert_p<2^{-2}$ for all $m\ge n_2$ and since $n_2>n_1\ge n(1)$ from the previous step results that $\lVert f_{n_2}-f_{n_1}\lVert _p<2^{-1}$. In this way we make an increasing sequence $n_k$ such that $$\lVert f_{n_{k+1}}-f_{n_k}\rVert_p<2^{-k}\quad \forall k\in\mathbb{N}.$$Question 1 Is this enough to justify what the author said?

Now, we define $$g_k= \lvert f_{n_1}\rvert+ \lvert f_{n_2}-f_{n_1}\rvert+\cdots+\lvert f_{n_{k+1}}-f_{n_k} \rvert,\quad \text{for}\ k=1,2,3,\dots$$.It is clear that from the Minkowski's inequality we have that\begin{eqnarray*}\lVert g_k^p \rVert_1=\lVert g_k\rVert_p^p&=&(\rVert\lvert f_{n_1}\rvert+ \lvert f_{n_2}-f_{n_1}\rvert+\cdots+\lvert f_{n_{k+1}}-f_{n_k} \rvert\rVert_p)^p\\&\le&\left ( \lVert f_{n_1}\rVert_p+\sum_{j=1}^k\lVert f_{n_{j+1}}-f_{n_j}\rVert_p\right)\le (\lVert f_{n_1}\rVert_p)+\alpha)^p<\infty\end{eqnarray*}

Let $g=\lim_{k\to\infty} g_k$. By B. Levi's theorem and the above estimate we have $$\int g^p\;d\mu=\int\lim_{k\to\infty}g_k^p\;d\mu=\lim_{k\to\infty}\int g_k^p\;d\mu<\infty.$$ Hence $g$ is in $L^p$; i.e $$\int\left [ \lvert f_{n_1}\rvert +\sum_{j=1}^\infty \lvert f_{n_{j+1}}-f_{n_j}\rvert\right]^p\;d\mu<\infty$$ The nonnegative integrand above must be finte a.e, and so the series $\lvert f_{n_1}(x)\rvert +\sum_{j=1}^\infty \lvert f_{n_{j+1}}(x)-f_{n_j}(x)\rvert$ converges a-e and therefore also the series $$f_{n_1}(x)+\sum_{j=1}^\infty(f_{n_{j+1}}(x)-f_{n_j}(x))$$ converges a.e. The $k$ th partial sum of this serie is $f_{n_{k+1}}(x)$ and so the sequence $(f_{n_k}(x))$ converges to a complex number $f(x)$ for all $x\in A$, where $A\in\mathcal{A}$ and $\mu(X\setminus A)=0$. Define $f(x)$ as $0$ for all $x\in X\setminus A$.

We will show that $f$ is the limit in $L^p$ of the sequence $(f_n)$, and this will of course prove that $L^p$ is complete in the metric induce by the $L^p-$norm. Given $\varepsilon > 0$, then from definition of Cauchy sequence exists $n(\varepsilon)\in \mathbb{N}$ such that $\lVert f_s -f_t\rVert_p <\varepsilon$ for all $s,t\ge n(\varepsilon)$, so let $l$ so large such that $$\lVert f_s - f_t\rVert_p<\varepsilon\quad\text{for}\quad s, t\ge n_l$$ (practically $n_l> n(\varepsilon)$, this is possible, we remembert that $\{n_k\}$ is increasing). Then for $k\ge l$ and $m>n_l$, we have $$\lVert f_m-f_{n_k}\rVert_p<\varepsilon.\tag1$$ By Fatou's lemma we have\begin{eqnarray*}\int \lvert f-f_m \rvert^p\;d\mu&=&\int\liminf_{k\to\infty}\lvert f_{n_k}-f_m\rvert^p\;d\mu\\&\le&\liminf_{k\to\infty}\lVert f_{n_k}-f_m\rVert_p^p\color{red}{\le} \varepsilon^p\end{eqnarray*}

Question 2. The inequality in red comes from $(1)$, but I don't know how to see it formally, could you explain it to me?

Thus for each $m>n_l$, the function $f-f_m$ is in $L^p$ and then chosen any $m>n_l$ we have that $f=(f-f_m)+f_m$ is in $L^p$.This implies that $f$ is in $L^p$ and we have done. Thanks for the attention.