Find $\Omega$ s.t. $L^1_{loc}(\Omega)$ is a normed space Is there any measurable $\Omega\subset\mathbb{R}^N$ for some $N\in\mathbb{N}$ with infinite Lebesgue-measure such that there exists a norm for $L^1_{loc}(\Omega)$?
 A: Let me clarify what Jochen means. Such set $\Omega$ you are searching for indeed exists. In fact every measurable set $\Omega \subseteq \mathbb{R}^N$ does the job. Due to Zorn's Lemma we know there exists a (Hamel) basis $\mathcal{B}$ of $L^1_{\text{loc}}(\Omega)$, means every element in $L^1_{\text{loc}}(\Omega)$ has a unique finite representation of elements in $\mathcal{B}$. So let us fix some $f \in L^1_{\text{loc}}(\Omega)$, then there exists an index $n \in \mathbb{N}$, elements $b_1, \dots b_n \in \mathcal{B}$ and coefficients $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ (assuming we are talking about $L^1_{\text{loc}}(\Omega)$ as a real vector space) with
\begin{align}
f = \sum_{i=1}^n \lambda_i b_i.
\end{align}
The expression
\begin{align}
\| f \| = \left( \sum_{i=1}^n \lambda_i^2 \right)^\frac{1}{2}
\end{align}
then gives rise to a norm on $L^1_{\text{loc}}(\Omega)$. Of course such a norm has no connection to the canonical Fréchet-topology on $L^1_{\text{loc}}(\Omega)$ and of course one can switch $L^1_{\text{loc}}(\Omega)$ for an arbitrary real or complex vector space $V$.
