I'm not sure what your sets are, but it seems you have the right idea. Here is an outline of one proof:
First prove the
Lemma: If $f_n\rightarrow f$ and $\int_E|f_n|\le M$ for each $n$, then $\int_E|f|\le M$. (Use Fatou.)
Having done that, proceed to prove the main result as follows:
Step 1) Use tightness to find a set $A$ of finite measure such that $\int_{A^C} |f_n|\le1$ for each $n$. Note by the Lemma then, that $\int_{A^C} |f|\le 1$.
Step 2) Use uniform integrability to find a $\delta>0$ such that whenever $\mu(E)<\delta$ then $\int_E |f_n|\le 1$ for each $n$. By the Lemma, then, we would have $\int_E |f|\le1$ for such $E$.
Step 3) Apply Egoroff to find $B\subset A$ with $\mu(A\setminus B)<\delta$ such that $f_n$ converges uniformly on $B$. Now pick $N$, so that for $m,n\ge N$, $\int_B|f_n-f_m|\le1$.
Step 4) From step 3) we have $\int_{ B}|f_n|\le 1+\int_X|f_N|$ for each $n\ge N$. Thus, there exists an $M$ so that $\int_{ B}|f_n|\le M$ for all $n$. Use the Lemma once more to conclude that $\int_{ B}|f|\le M$.
Step 5) Conclude that $\int_X |f| =\int_{A^C}|f|+\int_{A\setminus B}|f|+\int_B|f|<\infty$.