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I am a physics undergraduate so I only had two lectures about measure theory. Could someone explain to me what is the most correct statement between these two ($p<\infty$):

  1. From Is $L^p$ separable?. Let $(X,\sigma,\mu)$ be a measure space where $\mu$ is a sigma finite measure. If there exists a succession $A=A_1,...,A_n,... ⊂ \sigma$ such that, for any $>0$ and $M∈Σ$ one can find $A_k∈A$ with $(_k△M)<$ then the metric space $(L_p,||\cdot||_p)$ is separable.
  2. From Countably generated $\sigma$-algebra implies separability of $L^p$ spaces. Let $(X,\sigma,\mu)$ be a measure space where $\mu$ is a sigma finite measure. If the sigma algebra $\sigma$ is countably generated (i.e. we can find a succession $A=A_1,...A_n,...$ such that $\sigma$ is generated by $A$) then the metric space $(L_p,||\cdot||_p)$ is separable.
davise
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    What would you deem as "most correct"? Looking at the posts linked (and the links to said posts) it seems to me that both might be true but that both aren't necessary conditions for $L^p$ to be separable. It does sound like the first one can be turned into a necessary and sufficient criterion if it's only stated for $M$ of finite measure but I wouldn't know myself. Hopefully someone more knowledgeable passes by and answers you more confidently. – Bruno B Nov 28 '23 at 12:02
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    What do you mean by "most correct"? Both are correct. Result 2 implies result 1. Note that, if $(X,\sigma,\mu)$ is a measure space where $\mu$ is a sigma finite measure and there exists a succession $A=A_1,...,A_n,... ⊂ \sigma$ such that, for any $>0$ and $M∈Σ$ one can find $A_k∈A$ with $(_k△M)<$, then the sigma algebra $\sigma$ is countably generated. – Ramiro Nov 28 '23 at 12:07
  • Hi, thanks for your answer. I would say that if one of the two conditions implies the other one, it is the most correct. Also, I don't have the knowledge to understand which of the two statements is the most rigorous. I am only interested in a sufficient condition for Lp to be separable. – davise Nov 28 '23 at 12:24
  • @Ramiro This is what I was looking for. Based, on what you wrote I would say result 1 implies result 2. – davise Nov 28 '23 at 12:30
  • For most correct i mean the weaker condition for Lp to be separable. Reading @Ramiro wrote I think 1 is weaker than 2. (Probably this was very trivial but I barely studied the basics of measure theory). – davise Nov 28 '23 at 12:33
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    'Weaker' and 'stronger' is the most correct term to compare theorems. :) – daw Nov 28 '23 at 15:50
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    Looking more carefully, we can prove that: $(X,\sigma,\mu)$ is a measure space where $\mu$ is a sigma finite measure and there exists a succession $A=A_1,...,A_n,... ⊂ \sigma$ such that, for any $>0$ and $M∈\sigma$ one can find $A_k∈A$ with $(_k△M)<$ if and only if the sigma algebra $\sigma$ is countably generated. So result 1 and result 2 are equivalent. None of them is weaker than the other. – Ramiro Nov 29 '23 at 12:15

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