My question is about why weights are defined to be locally integrable.
I am trying to understand a statement in the paper "Poincare Inequalities and Neumann problems for the $p$-Laplacian" by David Cruz-Uribe, Scott Rodney and Emily Rosta. The paper can be found on the arXiv here.
Lemma 2.2 in the paper is where my question stems from. My question involves this particular statement in the proof: "since $\lambda_j^{p/2}(x) \leq \gamma(x) \in L^1_{loc}(E)$ a.e. the spaces $L^p(\lambda_j^{p/2};E)$ are complete..."
The proof that $L^p(\mathbb{R})$ is a banach space translates virtually unchanged into a proof that $L^p(w;\mathbb{R})$ is a banach space for a weight $w$. It doesn't seem to need the condition that $w \in L^1_{loc}(\mathbb{R})$.
I know weights are defined to be locally integrable, but is local integrability a necessary condition for a weighted Lebesgue space to be complete? Is there an example of a weight $w$ with $w \not\in L^1_{loc}(E)$ such that $L^p(w;E)$ is not complete?
Not, it is not needed, however, it is usually added for convenience as weights appearing in the real-life scenarios are usually locally integrable and this allows for using certain separation arguments. So what is a weight really?
Let $(X, \mu)$ be a measure space and let $w\colon X\to \mathbb{R}$ be a non-negative, measurable function. Note that in this setting we cannot talk about local integrability as we don't have any topology at hand.
As $w \geqslant 0$, the integral $\int_X w\, {\rm d}\mu$ exists but may be infinite. Then you may consider $L_p(w, \mu)$, which is $L_p(\nu)$ for $\nu(A) = \int_A w {\rm d}\mu$. So as such this space is complete. The moral of that story is that weighted $L_p$-spaces are still $L_p$-spaces (and as such complete) but under a different measure.