Saying $\lim\int|f-f_n|=0$ is not equivalent to saying $\lim\int f_n=\int f$ in general. The first implies the second (since $|\int f_n-\int f|=|\int(f_n-f)|\le\int|f_n-f|$) but not conversely.
Wondering where you could have got the idea that they were equivalent, I think I got it. The following two theorems are equivalent:
Thm 1. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $\int g<\infty$, and $|f_n|\le g$. Then $\int f_n\to\int f$.
Thm 2. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $\int g<\infty$, and $|f_n|\le g$. Then $\int|f_n-f|\to0$.
Note that saying the theorems are equivalent does not say that the conclusions of the theorems are equivalent.
It's trivial that Theorem 2 implies Theorem 1. Since I suspect that showing the two theorems are equivalent was a homework problem, I'll just give a huge hint how to show that Theorem 1 implies Theorem 2:
Suppose Theorem 1. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $|f_n|\le g$, and $\int g<\infty$. Let $F_n=???$ and let $G=???$. Then $F_n\to F$ almost everywhere, where $F=???$. Also $G\ge0$, $|F_n|\le G$, and $\int G<\infty$. So Theorem 1 implies that $\int F_n\to\int F$, and it follows that $\int|f_n-f|\to0$.
(That does not show that $\int f_n\to\int f$ and $\int|f_n-f|\to0$ are equivalent, because in deriving Theorem 2 from Theorem 1 we applied Theorem 1 to the sequence $(F_n)$, not to the seqence $(f_n)$.)