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Whether a $L^p(X,\mu)$ space is separable? I understand that the answer depends on $p$ and $X$.

It seems to me that it is separable when $1\leq p < \infty, X=\mathbb{R}^n$ or $X=\mathbb{N}$. Also, $L^\infty(X,\mu) $ is seperable iff $X$ has only finite points.

Is there any general results? For example, does the separability of $(X,\mu)$ implies the separability of $L^p(X,\mu)$ when $p<\infty$?

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  • $\begingroup$ Wikipedia en.wikipedia.org/wiki/Separable_space#Non-separable_spaces: The Lebesgue spaces Lp, over a separable measure space, are separable for any 1 ≤ p < ∞. $\endgroup$
    – Urgje
    Jun 27, 2014 at 22:30
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    $\begingroup$ If $p=\infty$, then $L_p(X,\Sigma,\mu)$ is separable iff $X$ consist of finite amount atoms. If $1\leq p<\infty$, then $L_p(X,\mu)$ is separable iff $(X,\Sigma,\mu)$ is separable. This means that there exists a countable family $\mathcal{F}\subset\Sigma$ such that, for any $\varepsilon>0$ and $A\in\Sigma$ one can find $B\in\mathcal{F}$ with $\mu(A\triangle B)<\varepsilon$. $\endgroup$
    – Norbert
    Jun 28, 2014 at 9:42
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    $\begingroup$ Thank you so much for your help! Norbert, your answer is very comprehensive! The definition of the separability of $(X,\Sigma,\mu)$ is quite new to me. Could you give me a reference of the proof for your latter statement? Thank you! $\endgroup$
    – Zach
    Jun 28, 2014 at 15:45
  • $\begingroup$ @Norbert : When you say "$X$ consist of finite amount atoms", does this means that $X$ is the union of a finite number of atoms (and not that the number of atoms is finite)? $\endgroup$
    – Watson
    Mar 5, 2016 at 18:56
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    $\begingroup$ @Watson, exactly. I don't think my answer is good enough, because it will take too long to write down the proofs, and I don't have a good reference for these facts. $\endgroup$
    – Norbert
    Mar 5, 2016 at 20:57

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