TITLE: Pointwise convergence implies $L^p$
QUESTION [7 upvotes]: Simply, why is it that convergence pointwise, $u_j \rightarrow u$, implies convergence in $L^p$ if $|u_j(x)| \le g(x)$ for some $g$ in $L_+^p$?

REPLY [7 votes]: Since $u_j \to u$ pointwise and $|u_j(x)| \leqslant g(x)$, there is
$$
|u(x)|=\lim_{j\to\infty}|u_j(x)|\leqslant g(x)
$$
Since $|u_j(x)-u(x)|^p\to 0$ a.e, and
$$
\int_A |u_j(x)-u(x)|^p\:d\mu\leqslant \int_A |u_j(x)|^p\:d\mu+\int_A |u(x)|^p\:d\mu\leqslant 2\int_A |g(x)|^p\:d\mu
$$
where $|g(x)|^p$ is integrable. 
By Lebesgue's dominant convergence theorem 
$$
\lim_{j\to\infty}\int_A |u_j(x)-u(x)|^p\:d\mu=0
$$