I found this exercise:
- Given $(X,\mu)$ space with measure, $(u_n) \subset L^1(X,\mu)$, $u_n \rightarrow u$ almost everywhere, $|u_n|\leq |u|$ almost everywhere, prove that $u_n \rightarrow u$ in $L^1$.
I think this is precisely the Lebesgue theorem, so applying it you obtain:
$lim_n \int_X u_n d\mu = \int_X u d\mu $
$lim_n \int_X (u_n-u) d\mu =0 $
$lim_n \int_X |u_n-u| d\mu =0 $
which is the definition for $u_n \rightarrow u$ in $L^1$.
Can it be the proof? Thanks.
Not really. For $\displaystyle\int_{X}(u_{n}-u)d\mu\rightarrow 0$ does not entail $\displaystyle\int_{X}|u_{n}-u|d\mu\rightarrow 0$, but the latter will imply the former. Rather, you may look at $|u_{n}-u|\leq|u_{n}|+|u|\leq 2|u|$ and apply Lebesgue Dominated Convergence Theorem to the whole $|u_{n}-u|$ to finish the job.