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?

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.
