Is $\lim_{p \searrow 1} \|u\|_{L^p(\Omega)} = \|u\|_{L^1(\Omega)}$ true? Let $\Omega \subset \mathbb{R}^n$ be a bounded, open set and $u \in L^q(\Omega)$ for some $q \gt 1$ (ie. $u \in L^p(\Omega)$ for all $p \in [1,q]$).
Is then
$$
\lim_{q \ge p \searrow 1} \left( \int_\Omega |u|^p \right)^\frac1p = \int_\Omega |u|
$$
true? (In more colloquial terms: Are the $L^p$ norms continuous in $p=1$?)
Sadly, I have no idea how to prove that. (I just guess it should be true.)
Furthermore, are there similar results for $p \in (1,\infty)$?

You need to assume $u\ge 0.$ For $1\le p \le q, u^p \le 1+u^q \in L^1(\Omega).$ (To see this, think about the sets where $0\le u \le 1, u> 1.$) So by the dominated convergence theorem,
$$\lim_{p\to 1^+}\int_{\Omega} u^p \to \int_{\Omega} u^1.$$
Because $1/p \to 1,$
$$\lim_{p\to 1^+}\left( \int_{\Omega} u^p\right)^{1/p} = \left(\int_{\Omega} u^1\right )^1.$$ 
