Show that the limit sequence $f$ is in $\mathcal{L}^1$ For $\Omega = \mathbb{N}$ with counting measure is $\mathcal{L}^1$ complete.
Let $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1$ such that  every $f_k$ is a sequence $(f^n_k)_{n \in \mathbb{N}}$.
I already showed that a sequence $f=(f^n)_{n \in \mathbb{N}}$ exists such that $f_k$ converges pointwise to $f$. How do I show that the limit sequence $f$ is in $\mathcal{L}^1$?
 A: Attention: Since the functions $f_k$ are functions defined on $\mathbb{N}$, it is clearer to write $f_k(n)$ and $f(n)$ instead of $f_k^n$ and $f^n$.
Let us prove the result.
Since $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1(\mathbb{N},\#)$ (where $\#$ is the counting measure), we have that, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, for all $r,s >M$,
$$ \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon/2$$
So, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, for all $r,s >M$ and for all $p\in\mathbb{N}$,
$$ \sum_{i=0}^p |f_r(i) -f_s(i)| \leq \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon/2$$
Now, consider a fixed $p\in\mathbb{N}$, since $f_r$ converges pointwise to $f$, we have, for all $s >M$
\begin{align*} 
\sum_{i =0}^p |f(i) -f_s(i)|&= \sum_{i =0}^p  | \lim_{r \to +\infty}f_r(i) -f_s(i)| =\\
&=\sum_{i =0}^p \lim_{r \to +\infty} |f_r(i) -f_s(i)|=\\
&=\lim_{r \to +\infty} \sum_{i =0}^p |f_r(i) -f_s(i)|\leq\varepsilon/2 
\end{align*}
So, for any $p\in\mathbb{N}$, for all $s >M$
$$\sum_{i =0}^p |f(i) -f_s(i)| \leq\varepsilon/2$$
So,  for all $s >M$,
$$ \int |f -f_s| d\#=\sum_{i \in \mathbb{N}} |f(i) -f_s(i)| = \sup_p \left(\sum_{i =0}^p |f(i) -f_s(i)|\right) \leq\varepsilon/2<\varepsilon \tag{1} $$
It also follows that, for all $s >M$,
$$\left | \int |f| d\# - \int |f_s| d\#\right | \leq \int | \, |f|-|f_s| \, | d\# \leq \int |f -f_s| d\# \leq\varepsilon/2<\varepsilon \tag{2}$$
So, by $(2)$ we see that $\int |f| d\# <+\infty$, so $f \in \mathcal{L}^1(\mathbb{N},\#)$.
By $(1)$, we have that $f_s$ converges to $f$ in $\mathcal{L}^1(\mathbb{N},\#)$.
And, by $(2)$ again, we see that $\int |f_s| d\#$ converges to $\int |f| d\#$. That means, $\|f_s\|_1$ converges to $\|f\|_1$.
