Sequence of $L_1$ functions converging ae ....

Question

$(f_n) \subset L_1[0,1]$ coverges a.e. to $g$ with properties $\|f_n\|_1 = 2$ and $\|g\|_1 =1$.

Prove that

$$ \lim_{n \to \infty} \int_0^1 |f_n - g| =1 $$

This is a problem from our practice analysis comp. Help please!

Answer 1

Since $\|f_n - g\|_1 \ge \|f_n\|_1 - \|g\|_1 = 1$, we get

$$\liminf_n \int_0^1 |f_n - g| \ge 1.$$

On the other hand, since $|g| + |f_n| - |f_n - g|$ is nonnegative and converges a.e. to $2|g|$, Fatou's lemma gives

$$\liminf_n \int_0^1 (|g| + |f_n| - |f_n - g|) \ge \int_0^1 2|g|.$$

Show that this implies

$$\limsup_n \int_0^1 |f_n - g| \le 1.$$