Is $L^p$($\Omega$) for $\Omega$ locally compact (in terms of radon measure) separable?

Question

I have read in my script, that for $\Omega$ locally compact and sigma-compact $L^p$($\Omega$) is separable in terms of radon measure.For $\Omega$ Lebesgue-measurable $L^p$($\Omega$) is separable too, but about locally compact I didnt find any answer.

Answer 1

$\Omega$ locally compact does not imply that $L^p(\Omega)$ is separable. Note that the discrete topological space on any set is locally compact. Take $\Omega$ to be an uncountable set equipped with the discrete topology and define a measure on $\mathcal{P}(\Omega)$ by setting $\mu(\{x\}) = 1$ for any $x \in \Omega$.

Then $L^p(\Omega, \mu)$ is not separable. Indeed if $f \in L^p(\Omega, \mu)$ then $f$ has countable support so for any countable set $\{f_j \in L^p(\Omega, \mu) \mid j \geq 1\}$, $\mathcal{A} = \bigcup_{j \geq 1} \mbox{supp}(f_j)$ is countable and in particular there exists $x \in \Omega \setminus \mathcal{A}$. Then $g(y) = \mathbb{1}_x(y) \in L^p(\Omega, \mu)$ and it is easily verified that $g$ is distance at least $1$ from each of the $f_j$s and hence $\{f_j \in L^p(\Omega, \mu) \mid j \geq 1\}$ is not dense.