$\sigma$-finiteness of measures and separability of $L^p$ spaces

Question

The fact that a measure $\mu$ is $\sigma$-finite determines or not the separability of $L^2(\mu)$? I proved that $L^2([0,1])$ endowed with the counting measure $m$ is not separable since it admits an uncountable orthonormal basis. Can I reach the same conclusion by using instead a $\sigma$-finite measure? I tried to reason with the Lebesgue measure but it hasn't lead to anything.

Answer 1

$\sigma$-addtivity is not enough. Separability of $L^p(\mu)$ $1\leq p<\infty$ depends and a finer structure of the $\sigma$-algebra $\mathcal{F}$ on which $\mu$ acts, namely countably generated. That is, if there is a countable collection of sets $\mathcal{C}$ such that $\sigma(\mathcal{C})=\mathcal{F}$, then any $L_p$ $1\leq p<\infty$ will be separable. This is the case because simple functions are dense in $L_p$.

Suppose $\mathcal{C}$ is a countable collection of subsets of $\Omega$ and $\mathscr{F}=\sigma(\mathcal{C})$. Let $\{C_n:n\in\mathbb{N}\}$ be an enumeration of the elements of $\mathcal{C}$. For each $n\in\mathbb{N}$ define $\mathcal{A}_n$ the algebra generated by $\{C_1,\ldots, C_n\}$. Then $\mathcal{A}=\bigcup_n\mathcal{A}_n$ is a countable algebra that contains $\mathcal{C}$ and $\sigma(\mathcal{A})=\mathscr{F}$.

For simplicity, assume $\mu(\Omega)<\infty$. Using the procedures of Lebegue-Caratheodory it is easy to check that for any $E\in\mathscr{F}$, there is a sequence $A_n\in\mathcal{A}$ such that $\mu(E\triangle A_n)=\|\mathbb{1}_E-\mathbb{1}_{A_n}\|^p_p\xrightarrow{n\rightarrow\infty}0$ ($1\leq p<\infty$). From this, it is easy to see that $L_p(\mu)$ is separable, for simple functions are dense in $L_p(\mu)$.