$L^p$ space is separable if and only if measure space separable I am currently dealing with the following question:

Let $(E,\mathcal{A},\mu)$ be measure space, $p\in[0,+\infty)$. Show that $L^{p}(\mu)$ has a countable dense set iff there exists $(A_n)\in \mathcal{A}^{\mathbb{N}}$ such that $\forall n\in \mathbb{N},\ \mu(A_n)&lt;\infty$ and $\forall \epsilon&gt;0,\ \forall A \in \mathcal{A}\ s.t.\ \mu(A)&lt;\infty, \exists n\in \mathbb{N}\ s.t. \mu(A_n\Delta A)\leq \epsilon$.

For the first part, to show $L^p$separable, I have a sketch of proof that I am not sure of: let $\mathcal{F} = \{ \text{finite sum of functions in the form } f_{n,k} = q_k \mathbb{1}_{A_n}\}$, where $q_k\in \mathbb{Q}$ and $A_n$ is a member of the countable dense set. Then $\mathcal{F}$ would be dense in the set of integrable simple functions. By that simple functions are dense, we could reach that $\mathcal{F}$ is dense in $L^p$. But for the converse way, I really have no idea.

Thanks!
If $L^p$ is separable let $G$ be a countable dense subset. For rational $q,q'$ with $q&lt;q'$ and $0\not \in [q,q']$ and for $g\in G$ let $H(g,q,q')=g^{-1}[q,q').$ Let $\{A_n: n\in \Bbb N\}$ consist of the empty set and  every $H(g,q,q').$ 

If $A\in \mathcal A$ with $0&lt;\mu(A)&lt;\infty,$ let $f=\chi_A$ be the characteristic (indicator) function of $A.$ Given $\epsilon &gt;0,$ take $g\in G$ with $\int |f-g|^pd\mu&lt;\epsilon^p \mu(A)$   and consider $H(g,1-r,1+r)$ where $r\in \Bbb Q^+$ is arbitrarily small.
If $L^p$ is separable let $G$ be a countable dense subset. For rational $q,q'$ with $q&lt;q'$ and $0\not \in [q,q']$ and for $g\in G$ let $H(g,q,q')=g^{-1}[q,q').$ Let $\{A_n: n\in \Bbb N\}$ consist of the empty set and  every $H(g,q,q').$ 

If $A\in \mathcal A$ with $0&lt;\mu(A)&lt;\infty,$ let $f=\chi_A$ be the characteristic (indicator) function of $A.$ Given $\epsilon &gt;0,$ take $g\in G$ with $\int |f-g|^pd\mu&lt;\epsilon^p \mu(A)$   and consider $H(g,1-r,1+r)$ where $r\in \Bbb Q^+$ is arbitrarily small.