$L^\infty$ with the counting measure on the sigma-algebra of countable and co-countable sets. I am reading folland chapter 6.2 (The Dual of $L^\infty$). Here is a paragraph of this chapter on page 191,

Let $X$ be an uncountable set, $\mu=$ counting measure on $(X, \mathcal{P}(X)), \mathcal{M}=$ the $\sigma$ algebra of countable or co-countable sets, and $\mu_{0}=$ the restriction of $\mu$ to $\mathcal{M}$. Every $f \in L^{1}(\mu)$ vanishes outside a countable set, and it follows that $L^{1}(\mu)=L^{1}\left(\mu_{0}\right)$. On the other hand, $L^{\infty}(\mu)$ consists of all bounded functions on $X$, whereas $L^{\infty}\left(\mu_{0}\right)$ consists of those bounded functions that are constant except on a countable set. With this in mind, it is easy to see that the dual of $L^{1}\left(\mu_{0}\right)$ is $L^{\infty}(\mu)$ and not the smaller space $L^{\infty}\left(\mu_{0}\right)$.

I don't understand why $L^{\infty}\left(\mu_{0}\right)$ consists of those bounded functions that are constant except on a countable set. Can someone give me a hint? Thanks!
 A: Hint:
It is enough to assume that $f\geq0$ and not identically $0$.
Recall that
$$\|f\|_{L_\infty(\mu_0)}=\inf\{\alpha>0: \mu_0(f>\alpha)=0\}$$
In particular, it follows that $m(|f|>\|f\|_{L_{\mu_0}})=0$. This means that $\{f>\|f\|_{L_\infty(\mu)}\}=\emptyset$.  Without loss of generality, assume that $\|f\|_{L_\infty(\mu_0)}=1$ (divide $f$ by $\|f\|_{L_{\infty}(\mu_0)}$)
The sequence
$$s_n=\sum^{2^n-1}_{k=0}k2^{-n}\mathbb{1}_{\{k2^{-n}\leq f<(k+1)2^{-n}\}}+\mathbb{1}_{\{f=1\}}$$
is monotone nondecreasing (in $n$) and converges pointwise to $f$.
Each $s_n$ is $\mathcal{M}$-measurable. For each $n\in\mathbb{N}$, define $A_{nk}=\{k2^{-n}\leq f<(k+1)2^{-n}\}$, $k=1,\ldots,2^{n}-1$ and $A_{n,2^n}=\{f=1\}$. The sets $A_{n,k}$, $0\leq k\leq 2^n$ are pairwise disjoint; hence all,  but exactly one of them,  are countable. For each $n$, let $A_{n,k_n}$ be the unique uncountable set. Notice that $A_{n+1,k_{n+1}}\subset A_{n,k_n}$ and $\operatorname{diam}(A_{n,k_n})\xrightarrow{n\rightarrow\infty}0$.
The density of the dyadic numbers imply that there is one number $0\leq\alpha\leq1$ such that $\{f\neq\alpha\}$ is countable and so, $\{f=\alpha\}$ is countable. The conclusion will follow from this.
A: Let
$$\alpha = \inf_a \{a \mid \{f(x)<a \text{ is uncountable}\}\}$$
Since $f$ is bounded $\{f(x)<a\}$ is empty for some $a$ and the whole (uncountable) domain for some other $a$ . Therefore $\alpha$ is finite.
For every $\beta > \alpha$, $\{f(x) < \beta\}$ is uncountable and hence $\{f(x) \ge \beta\}$ is countable.
For every $\beta<\alpha$, $\{f(x) < \beta\}$ is countable.
Therefore,
$$\{f(x) \ne \alpha\}=\left(\bigcup_{n=1}^\infty \{f(x) < \alpha-1/n\}\right) \bigcup \left(\bigcup_{n=1}^\infty \{f(x) \ge \alpha+1/n\}\right), $$ is countable. That is $f(x)=\alpha$ except on a countable set.
