counting measure space is not separable but the corresponding $L^p$ space is separable Let $\left(X,\mathcal{F},\mu\right)$ be a measure space. We define a pseudometric on measure space: for any $A,B\in\mathcal{F}$ $$d_{\mu}\left(A,B\right)=\mu\left(A\Delta B\right)=\mu\left(\left(A\setminus B\right)\cup\left(B\setminus A\right)\right)$$
 $d_{\mu}$ becomes a metric if $\mathcal{F}$ is considered the equivalence relation $X\sim Y$ if and only if $d_{\mu}\left(X,Y\right)=0$ for any $X,Y\in\mathcal{F}$. The measure space $\left(X,\mathcal{F},\mu\right)$ is called separable measure space iff $\mathcal{F}$ is separable with respect to metric $d_{\mu}$. In other words, there exists a countable sequence $S$ of measurable sets in $\mathcal{F}$ , such that for all measurable sets $A\in\mathcal{F}$ and for all $\epsilon>0$ there exists $B\in S$ such that $d_{\mu}\left(A,B\right)<\epsilon$.
We have the conclusion that space $L^{p}\left(X,\mathcal{F},\mu\right)\left(1\le p<+\infty\right)$ is separable iff $\left(\left(X,\mathcal{F},\mu\right),d_{\mu}\right)$ is separable.
See this Space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable.
Consider counting measure space, $X=\left\{ 1,2,\dots\right\}$  ,$\mathcal{F}=2^{X}$, $\mu\left(\left\{ x\right\} \right)=1$ for $x\in X$, we can see that for any $A,B\in\mathcal{F}$ and $A\ne B$, we have $d_{\mu}\left(A,B\right)\geq1$ and we know $\mathcal{F}$ is uncountable, so counting measure space $\left(X,\mathcal{F},\mu\right)$ is not separable. 
On the other hand, as $L^{p}\left(X,\mathcal{F},\mu\right)$ is $l^{p}$ space , and $l^{p}$ is separable as $X$ is countable, so $\left(\left(X,\mathcal{F},\mu\right),d_{\mu}\right) $should be separable.
Can someone figure what's wrong here?
 A: Here is what is wrong: the correct statement is that the separability of $L^p$, $1\le p <\infty$, is equivalent to the existence of a countable collection $C$ of measurable sets of finite measure such that for each measurable set $Q$ of finite measure and for each positive $r$ there exists a set $A\in C$ such that the measure of $Q\Delta A$ is less that $r$. Hence, it is not the collection of all measurable sets but only the collection of measurable sets of finite measure that needs to be considered in the statement of separability on the side of sets.
The paper Separability of the L1 space of a vector measure by Werner J. Ricker, published in the Glasgow Mathematical Journal in 1992 refers to this as a ''classical result'' and points to the book of A. C. Zaanen ''Integration'' published in 1967 for a proof. The proof given in this book is not convincing, but the result is true and I can provide a proof if anyone is interested.
A: The counting measure is separable.  A measure space is separable iff it is generated by a countable collection of sets, modulo completion.   In this case, the singletons generate the $\sigma$-algebra that is the power set.
