$\{f_n\}_n$ converges to $f$ in $L^1$ means:
- $f_n \in L^1$ for every $n \in \mathbb{N}$;
- $\lim_{n \to +\infty} \int |f_n-f| =0$.
$\lim_{n \to +\infty} \int |f_n-f| =0$, so the sequence of $\int |f_n-f|$ can at most have finite infinity, thus bounded after these finite entries. So by the definition, it also implies $f \in L^1$ since
$$ \int |f| d\mu = \int |f-f_n + f_n| d\mu \leq \int |f-f_n| d\mu + \int |f_n| d\mu$$
which is finite.
However, I encountered a question proving uniformly convergence imply convergence in $L^1$,
Suppose $f \in L^1$, $f_n$ uniformly converge to $f$, $\mu(\Omega) < \infty$, we need to prove $f_n$ converges to $f$ in $L^1$ .
First, I want to show $f_n \in L^1$
$$ \int |f_n| d\mu = \int |f_n-f + f| d\mu \leq \int |f-f_n| d\mu + \int |f| d\mu \leq \mu(\Omega) \sup_{x\in \Omega}{|f-f_n|} +\int |f| d\mu$$.
My question is even if we know $\{\sup_{x\in \Omega}{|f-f_n|}\}$ converge to zero, I can not say all of the entries of it are finite. Am I right? so it is not enough to show $f_n \in L^1$ for every $n \in \mathbb{N}$?
As $$ \int_\Omega |f_n-f|d\mu=\|f_n-f\|_1\le \|f_n-f\|_\infty =\text{ess-sup}|f_n-f|\le \sup_{x\in \Omega}{|f-f_n|},$$
you can see that $$\lim_{n \to +\infty} \sup_{x\in \Omega}|f-f_n|=0$$
implies
$$\lim_{n \to +\infty} \int |f_n-f| d\mu=0.$$
For the example you are seeking for, if I understood correctly, $$f_n(x)=\frac{1}{n}+\frac{1}{x}$$ converges to $$f(x)=\frac{1}{x},$$ uniformly, while both of them are unbounded.