Let
$$u_n = \max\{ f, f_n\} \quad \text{and} \quad l_n = \min \{ f, f_n \} $$
so that both $(u_n)$ and $(l_n)$ converge pointwise to $f$, $l_n \leq f \leq u_n$ and $|f - f_n| = u_n - l_n$. By DCT, it is clear that $\int l_n \, d\mu \to \int f \, d\mu$. thus from
$$ \int u_n \, d\mu = \int (f + f_n - l_n) \, d\mu ,$$
taking $n\to\infty$, we have $\int u_n \, d\mu \to \int f \, d\mu$. (Here the assumption that you are concerning is used.) Therefore we have
$$ \int |f - f_n| \, d\mu = \int u_n \, d\mu - \int l_n \, d\mu \to 0. $$
The monotone convergence theorem handles infinities gracefully, which can only be done for functions that are positive (or otherwise reasonably controlled from below). In particular, nowhere does it assume that $f_n$, $f$, or the integrals, are finite. This seems to be beyond the scope of Beppo Levi, so I'm not sure that fixing this issue alone is considerably easier than proving everything from scratch. But let me try.
Depending on your version of definitions, it may or may not be trivial that for a positive function $f$ the following special case of monotone convergence holds:
$$\intop_E f dm = \lim_{C \to +\infty, E_n \uparrow E} \intop_{E_n} \min(f, C) dm$$
where $E_n$ are sets of finite measure that approximate $E$ (I assume $\sigma$-finiteness; if it fails then we should restrict to $\{f > 0\}$; if it fails even there then monotone convergence holds almost trivially with both sides infinite).
Now in order to make use of Beppo Levi we should make the limit finite. I would do that by replacing $f_n$ by $f_{n,C,k} := (f_n \wedge C) \mathsf{1}[E_k]$ and $f$ by $f_{C,k} := (f \wedge C) \mathsf{1}[E_k]$ for some fixed $k$. Now we can safely apply Beppo Levi to the successive differences $f_{n+1,C,k} - f_{n,C,k}$ to obtain
$$\intop f_{C,k} dm = \lim_{n \to \infty} \intop f_{n,C,k} dm$$
Now take $C$ and $k$ to $\infty$ and interchange the limits. You can always do this with monotonely increasing limits (this is equivalent to rearrangement of terms in positive series, or Fubini on $\mathbb{N} \times \mathbb{N}$, or whatever you prefer). On the other hand, monotone convergence itself is about rearrangement or Fubini on $E \times \mathbb{N}$, so I'm not even sure you would view the things that I rely on as more basic than those that you prove...
Best Answer
Define $$g_n=f_{n}-f_1$$
Then $g_n$ is an increasing sequence of non-negative integrable functions, and $g_n\to g:=f-f_1$, thus by monotone convergence theorem $$\int g_n d\mu\to\int g d\mu$$
Thus the result follows.