In order to show tightness, fix $\varepsilon>0$. Then you get $N=N(\varepsilon)$ such that $\int_Xf_nd\mu\leq\varepsilon$ if $n\geq N+1$. Now, for all $n\leq N$, you can find a positive $M$ such that $\int_{\{f_n\geq M_n\}}f_nd\mu\leq \varepsilon$, using integrability of $f_n$. (if $f$ is integrable apply the monotone convergence theorem to $f\mathbf 1_{\{|f|\geq n\}})$
Put $A_n:=\{f_n\leq M_n\}$, then $A_n$ is measurable. Take $X_0:=\bigcap_{k=1}^NA_k^c$. Each $A_k^c$ has finite measure (since $\mu(A_k^c)\leq \frac 1{M_k}\int f_kd\mu$) so $X_0$ is of finite measure. Check that we have the wanted inequality.
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
Consider $E=(-3,3)$, and for $n=1,2, \ldots$ define the functions \begin{align}h_n(x) := \begin{cases} -2n - n^2x, \; \text{ if } x \in \left[-\frac{2}{n}, \,-\frac{1}{n} \right] , \\ n^2x , \; \text{ if } x \in \left[-\frac{1}{n} , \, \,\frac{1}{n}\right] , \\ 2n - n^2x , \; \text{ if } x \in \left[\frac{1}{n} , \,\frac{2}{n}\right], \\ 0, \; \text{ if } x \in \left(-3, -\frac{2}{n}\right) \cup \left(\frac{2}{n}, 3 \right). \end{cases} \end{align}
Notice $\lim\limits_{n\to \infty} h_n(x)=0$ and $\lim\limits_{n\to \infty}\int\limits_{E} h_n dm=0$. Let $\delta>0$ be given. By the Archimedean Property, we may find a positive integer $N$ such that $N \delta > 4$. So we have that $m\left( \left[-\frac{2}{N}, \frac{2}{N} \right]\right)< \delta$ but notice $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_n \right| dm=2$ whenever $n \geq N$*. Hence the sequence of functions $\{h_n\}_{n=1}^\infty$ is not uniformly integrable.
*Note: Technically it suffices to observe that $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_N \right| dm=2$ since $\neg \left(\forall \varepsilon\left(\varepsilon>0 \implies \exists \delta \left( \delta>0 \land \forall n \forall A \left(m(A)<\delta \implies \int_A |\,h_n| dm < \varepsilon \right)\right) \right) \right)$
$\iff \exists \varepsilon \left( \varepsilon>0 \land \forall \delta \left( \delta>0 \implies \exists N \exists A \left(m(A)<\delta \land \int_A |\,h_N| dm \geq \varepsilon\right)\right) \right)$.