Let $(X, \mathcal{M}, \mu)$ be a measure space. Let $f \in \mathcal{L}^1(X)$ and $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions in $\mathcal{L}^1(X)$ such that
$$\int_X f \,d\mu = \lim_{n \rightarrow \infty} \int_X f_n \,d\mu$$
Does at this point already follow that
$$\int_X |f| \,d\mu = \lim_{n \rightarrow \infty} \int_X |f_n| \,d\mu$$
I think this is true, but I struggle to prove this. It seems like I'm missing an obvious inequality.
Attempt:
$$\lim_{n \rightarrow \infty} \int_X |f| - |f_n| \,d\mu \leq \int_X |f - f_n| \,d\mu$$
At this point we can't be sure, that $f_n \stackrel{n \rightarrow \infty}{\longrightarrow} f$ in $L^1$, so this doesn't work. Is it possible the implication is just not true in general? I'd like to see a counterexample in this case.
I just figured it out by myself. The statement as given is not true. As an example consider $$f \equiv 0 \hspace{30pt} \text{and} \hspace{30pt}f_n(x) = \mathbb{1}_{(0, 1]}(x) - \mathbb{1}_{(1, 2]}(x)$$
Then we have
$$\int_\mathbb{R} f \,d\mu = 0 = \lim_{n \rightarrow \infty} 0 = \lim_{n \rightarrow \infty} \int_\mathbb{R} f_n \,d\mu$$
but
$$\int_{\mathbb{R}} |f| \,d\mu = 0 \neq 2 = \lim_{n \rightarrow \infty} \int_{\mathbb{R}} |f_n| \,d\mu$$