Meaning of convergence in $L^1$ norm
Not unless you add the condition that the sequence $f_n$ is in $L^1.$ If so, then for all $n\in \mathbb{N}$
$$ \int |f| d\mu = \int |f-f_n + f_n| d\mu \leq \int |f-f_n| d\mu + \int |f_n| d\mu$$
which is finite.
If you do not require $f_n$ to be in $L^1$ then we can easily find a sequence such that $ \int |f-f_n| d\mu \to 0 $ but $f \notin L^1.$ Just pick your favorite non-integrable function $g$ and make $f_n = f = g.$
The answer of Ragib Zaman is correct. However, I think that a sentence like "$\{f_n\}_n$ converges to $f$ in $L^1$" means:
- $f_n \in L^1$ for every $n \in \mathbb{N}$;
- $\lim_{n \to +\infty} \int |f_n-f| =0$.
This is a reasonable approach, since, in any (say) metric space $X$, $x_n \to x$ is rather immaterial if $x_n \notin X$. The case of a sequence $u_n = f_n -f \in L^1$ with $f_n \notin L^1$ is really a trap for homework :-)