Suppose that $\{f_n\}$ is a sequence complex measurable functions on a measurable space $(X,\Omega,\mu)$. Let $f$ be the pointwise limit of $f_n$. Does (1) implies (2)? where $$\lim_{n\to\infty}\int_Xf_n\,d\mu=\int_Xf\,d\mu.\tag{1}$$ $$\lim_{n\to\infty}\int_X|f_n|\,d\mu=\int_X|f|\,d\mu.\tag{2}$$
[EDIT] I don't know if the above statement is valid or not. So please give me a counterexample if it is false.
My trial: (1) implies $$\lim_{n\to\infty}\int_X(f_n-f)\,d\mu=0.\tag{3}$$ $$\lim_{n\to\infty}\int_X|f_n-f|\,d\mu=0.\tag{4}$$ $$\lim_{n\to\infty}\int_X\Big||f_n|-|f|\Big|\,d\mu=0.\tag{5}$$ $$\lim_{n\to\infty}\int_X|f_n|-|f|\,d\mu=0.\tag{6}$$ So (2) follows. I'm not certain whether (3) implies (4). That is to say that (7) implies (8) where $g_n\to0$ and $$\lim_{n\to\infty}\int_Xg_n\,d\mu=0.\tag{7}$$ $$\lim_{n\to\infty}\int_X|g_n|\,d\mu=0.\tag{8}$$ Let $A_n=\{x\in X:|u_n(x)|\ge |v_n(x)|\}$ and $B_n=X\setminus A_n$. Then $$\int_X|g_n|\,d\mu =\int_X\sqrt{(u_n(x))^2+(v_n(x))^2}\,d\mu \le\int_{A_n}\sqrt2|u_n|\,d\mu+\int_{B_n}\sqrt2|v_n|\,d\mu \le\sqrt2\int_X|u_n|\,d\mu+\sqrt2\int_X|v_n|\,d\mu$$ So, it is enough to consider "(7) implies (8)" for real $g_n$.
False: Let $X=\mathbb R$ with the lebesgue measure $m$. Let $$f_n(x)=\chi_{[-n,n]}\times\frac1{n^2}x.$$ Then $f_n\to0$ uniformly and $$\int_{\mathbb R}f_n\,dm=0$$ for every $n\in\mathbb N$. But $$\int_{\mathbb R}|f_n|\,dm=1$$ for every $n\in\mathbb N$.