# 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 statement "Space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable" is false. In fact, it is not true that "If space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable then $(\Sigma, \rho_\Delta)$ is separable". You have just showed a counter-exemple. May 10, 2018 at 12:49
• Yeah, I just figure it out...If $\left(\left(X,\mathcal{F},\mu\right),d_{\mu}\right)$ is separable, then $L^{p}\left(X,\mathcal{F},\mu\right)\left(1\le p<+\infty\right)$ is separable, but the reverse may not hold. Thanks! May 10, 2018 at 13:07
• Separability is a topological property, how can you conclude that $X$ is not separable after saying that there are uncoutablely many measurable sets? In fact, $X$ is separable, just because it is countable. And this is not a proper counter-example.@Ramiro @Leoalan.Huang
– Bach
Feb 21, 2020 at 22:39

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 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.
• However, I define $X$ to be countable here, may be the natural number set May 10, 2018 at 12:29