Suppose $f_n\rightarrow f$ pointwise a.e. Show that $f_n\rightarrow f$ in $L^1$ if and only if $\lvert\lvert f_n\lvert\lvert_1 \rightarrow \lvert \lvert f\lvert\lvert_1$.
What happens if it’s all in $L^2$?
So we prove two direction, first we have $\lvert f_n-f\lvert\rightarrow 0 $ because of the pointwise convergence, then by triangle inequality, we have: $\lvert\lvert f_n\lvert\lvert_1 - \lvert \lvert f\lvert\lvert_1\leq \lvert\lvert f_n-f\lvert\lvert_1=\int\lvert f_n-f\lvert\rightarrow0$.
But I have trouble proving the other direction? Any ideas? Thanks.
A useful tool for this is the generalised Lebesgue dominated convergence theorem, which allows the dominating function to be dependent on $n$.
Suppose $\|f_n\|_1 \to \|f\|_1$.
Let $g_n(x) = |f_n(x)| + |f(x)|$, $g=2|f|$, then $g_n(x) \to g(x)$ for ae. $x$, $\|g_n\|_1 = \|f_n\|_1+\|f\|_1$, $\|g\|_1 = 2\|f\|_1$ and $\|g_n\|_1 \to \|g\|_1$.
We have $|f_n(x)-f(x)| \le g_n(x)$ and since $f_n(x) \to f(x)$ for ae. $x$ we see that $\|f_n-f\|_1\to 0$.
This generalises in a straightforward manner if we note that $|a+b|^p \le 2^{p-1}(|a|^p+|b|^p)$.