$L^1([0,1])$ spaces

Question

Let $\left\{ { f }_{ n } \right\}$ be a sequence of functions in ${ L }^{ 1 }\left( \left[ 0,1 \right] \right) $ such that $\left\{ { f }_{ n } \right\}{ \longrightarrow } f$ pointwise, almost everywhere.Show that $\left\{ { f }_{ n } \right\} \longrightarrow f$ in ${ L }^{ 1 }\left( \left[ 0,1 \right] \right) $ if and only if $\left\{ { \left\| { f }_{ n } \right\| }_{ 1 } \right\} \longrightarrow { \left\| f \right\| }_{ 1 }$ in $\mathbb{R}$

Answer 1

One direction is easy. For the other direction, you need a general version of Dominated Convergence Theorem saying the following: if $f_n,f$ are in $L^1$ and $g_n,g \in L^1$ are non-negative such that $|f_n| \le g_n$ and $|f| \le g$ a.e., $f_n \to f$ a.e., and $\int g_n \to \int g$ then $f_n \to f$ in $L^1$.

To prove this generalization, just adapt the proof of dominated convergence theorem.

Answer 2

Let us prove the non trivial part of the result.

Since $|f_n - f| \leq |f_n| + |f|$, it is enough to apply Fatou's Lemma to the sequence of non-negative functions $g_n := |f_n| + |f| - |f_n - f|$, obtaining $$ \int_0^1 \liminf_n g_n = \int_0^1 2 |f| \leq \liminf_n \int_0^1 g_n = \int_0^1 2|f| - \limsup_n \int_0^1 |f_n-f|. $$ This gives $$ \limsup_n \int_0^1 |f_n-f|\leq 0, $$ i.e. $f_n \to f$ in $L^1$.