let $X$ be a finite measure space and $\{f_n\}$ be a sequence of integrable functions, $f_n \rightarrow f\text{ a.e.}$ on $ X$. I want to show if (1) holds, then (2) holds too.
$$\lim_{n \rightarrow \infty}\int_X |f_n| \, d\mu=\int_X |f| \, d\mu,\tag{1}$$
$$\lim_{n \rightarrow \infty}\int_X |f_n-f| \, d\mu=0.\tag{2}$$
My attempt:
I have proven that (2) holds for nonnegative $f$. Then for the general case, I split the set to $E^+=\{x: f \geq 0\}$ and $E^-=\{x: f \leq 0\}$:
$$\lim_{n \rightarrow \infty}\int_{E^+} f_n \, d\mu-\int_{E^+} f \, d\mu -\lim_{n \rightarrow \infty}\int_{E^-} f_n \, d\mu+\int_{E^-} f \, d\mu=0$$
But I don't know how to proceed from here!
As Lukas Geyer points out, the result isn't true unless $f$ is integrable. To see why, consider $f_n(x) = n$. Clearly, $f(x) = \infty$. Also: $$ \int_X f_n \, d\mu = n \mu(X) $$
Thus: $$ \lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu = \infty $$
Yet: $$ \lim_{n \to \infty} \int_X |f_n - f| \, d\mu = \infty \neq 0 $$
Now, assuming $f$ is integrable, we have: $$ \left||f_n - f| - |f_n|\right| \le |f| $$
Hence, by the dominated convergence theorem: $$ \lim_{n \to \infty} \int_X \left(|f_n - f| - |f_n|\right) \, d\mu = - \int_X |f| \, d\mu $$
Rearrange to get the required result.