I am working on a problem$^{(*)}$ on Lebesgue integral looks like this:
Given that both $f_n$ and $f$ are integrable, $f_n \longrightarrow f$ a.e., and $\int|f_n| \longrightarrow \int |f|$. Show that $$\int |f_n - f| \longrightarrow 0.$$
To me, the question "makes sense" since $\int |f_n - f| \longrightarrow 0$ implies $\lim_{n \to \infty} \int |f_n - f| = 0,$ and $|f_n - f|$ is approaching zero as $n$ is approaching infinity, therefore the integral is approaching zero. But I do not know how to say it mathematically.
Any help or hints would be very much appreciated. Thanks for your time.
(*) Richard F. Bass' Real Analysis, 2nd. edition, chapter 7: Limit Theorems, exercise 7.5, page 57.
Let $g_n = |f_n| + |f| - |f_n - f|$. Then $g_n \ge 0$ and $g_n \to 2|f|$ a.e., hence
$$\liminf_{n\to \infty} \int g_n \ge \int 2|f|$$
by Fatou's lemma. Since $\int |f_n| \to \int |f|$,
$$\liminf_{n\to \infty} \int g_n = \liminf_{n\to \infty} \int (|f_n| + |f|) - \limsup_{n\to \infty} \int |f_n - f| = \int 2|f| - \limsup_{n\to \infty} \int |f_n - f|.$$
Thus
$$\int 2|f| - \limsup_{n\to \infty} \int |f_n - f| \ge \int 2|f|.$$
Deduce from this inequality and the fact $\int 2|f| = 2\int |f| < \infty$ that
$$\int |f_n - f| \to 0.$$