I'm studying the proof of this Lemma.
Lemma
For $1\leqq p<\infty$ and $\{ f_k \}_{k=1}^\infty \subset L^p$, $$\sum_k \| f_k \|_p <\infty \Rightarrow \exists G\in L^p \ ; \ G(x)=\sum_k |f_k (x)|. $$
I have the point I don't understand.
Proof
Let $\{ f_k \} \subset L^p$ and suppose $\sum_k \| f_k \|_p <\infty$.
Define $G_n (x)=\sum_{k=1}^n |f_k (x)|$ for $n\in \mathbb N$.
For all $n$, $\| G_n \|_p\leqq \sum_{k=1}^n \| f_k \|_p \leqq \sum_k \| f_k \|_p$ thus $\{ G_n \}_{n=1}^\infty$ is bounded and monotone so converges to $\sum_k |f_k|=:G. \ (\sum_k |f_k(x)|=G(x) \ \mathrm{for \ all}\ x.)$
So I have to check $G\in L^p$. I don't know why this holds.
Now, $\left(\int |G(x)|^p d \mu \right)^{1/p} =\left(\int (\sum_k |f_k(x)|)^p d \mu \right)^{1/p} \cdots (\ast)$.
But why can we say $(\ast)$ is finite ? Each $f_k$ belongs to $L^p$ so $\left( \int |f_k(x)|^p d\mu \right)^{1/p}<\infty$, is this fact useful for $(\ast)<\infty$ ?
Let me write the proof of you statement from scratch filling some gaps that might cause confusion in first glimpse.
Let $G_n(x) = \sum_{k=1}^n|f(x)|$ for every $n\geq 1$. Then, for fixed $x$, $\{G_n(x)\}_{n=1}^{\infty}$ is increasing, and hence it converges to either a real number or tends to $+\infty$. Now, for every $n$ we have that
\begin{align} \biggl(\int G_n(x)^p\,d\mu\biggr)^{1/p}&=\biggl(\int \biggl(\sum_{k=1}^n|f(x)|\biggr)^p\,d\mu\biggr)^{1/p}\\ &\overset{\text{triangle ineq.}}{\leq}\sum_{k=1}^n\biggl(\int|f(x)|^p\,d\mu\biggr)^{1/p}\\ &\leq \sum_{k=1}^{\infty}\|f\|_p\,d\mu=:M<\infty.\tag{1} \end{align}
Now, since $G_n(x)^p$ is increasing with respect to $n$ it follows by the monotone convergence theorem that
\begin{align} \int \biggl(\sum_{k=1}^{\infty}|f(x)|\biggr)^p\,d\mu&=\int\lim_{n\to \infty}G_n(x)^p\,d\mu\\ &= \lim_{n\to \infty}\int G_n(x)^p\,d\mu\overset{(1)}{\leq}M^p<\infty\tag{2}. \end{align}
Therefore, it follows that $\biggl(\int |G(x)|^p\,d\mu\biggr)\leq M<\infty$, where $G(x):= \sum_{k=1}^{\infty}|f(x)|$. Hence, $G(x)<\infty$ almost everywhere, which means that $G(x)$ is well defined and by $(2)$ it belongs to $L^p$.