The zero-countable / infinity-cocountable measure space breaks the $p=1$ endpoint of duality

The family $\mathcal{A}$ contains $\mathrm{\varnothing }$ and is closed under complements by definition. Let $(E_n)$ lie in $\mathcal{A}$. If every $E_n$ is at most countable, then ${\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable by [L1]. If some $E_j$ is cocountable, then $X\setminus {\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\subseteq X\setminus E_j,$ so [L1] makes the complement at most countable. Thus $\mathcal{A}$ is a sigma-algebra on $X$.

The function $\mu$ satisfies $\mu (\mathrm{\varnothing })=0$. Let $(E_n)$ be pairwise disjoint in $\mathcal{A}$. If two members $E_i,E_j$ were cocountable, then $X=(X\setminus E_i)\cup (X\setminus E_j)$ would be at most countable by [L1], contradicting that $X$ is uncountable. So at most one $E_n$ is cocountable. If every $E_n$ is at most countable, then ${\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable by [L1], and both sides of countable additivity are $0$. If some $E_j$ is cocountable, then $X\setminus {\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\subseteq X\setminus E_j$ is at most countable by [L1], so the union is cocountable and both sides of countable additivity are $+\mathrm{\infty }$. Hence $\mu$ is a measure on $(X,\mathcal{A})$.

By definition of $\mu$, a measurable set has finite $\mu$-measure exactly when it is at most countable, and then its $\mu$-measure is $0$. In particular, $X$ is measurable and $\mu (X)=+\mathrm{\infty }$ because $X\setminus X=\mathrm{\varnothing }$ is at most countable.

Let $[f]\in L^1(\mu )$ and choose an integrable representative $u$. For each $n\ge 1$, set $E_n:=\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}n\}$. By [L4], $\mu (E_n)\le n\int \mathrm{\mid }u\mathrm{\mid }d\mu <\mathrm{\infty }\mathrm{.}$ Step 3.1 therefore gives $\mu (E_n)=0$, so each $E_n$ is at most countable. By [L1], $\{u\mathrm{\ne }0\}={\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable and hence null. Thus every element of $L^1(\mu )$ is the zero class.

The constant functions $0$ and $1$ are both in $L^{\mathrm{\infty }}(\mu )$, but they do not define the same $L^{\mathrm{\infty }}$ class because $\{1\mathrm{\ne }0\}=X,$ and step 3.1 gives that $X$ is not null. Thus $L^{\mathrm{\infty }}(\mu )$ contains at least two distinct classes.

Since $L^1(\mu )$ has only the zero class by step 4.1, its only bounded linear functional is the zero functional. Both $0$ and $1$ represent that zero functional, because $\int f\cdot 0d\mu =0=\int f\cdot 1d\mu$ for the unique class $f=0$ in $L^1(\mu )$. By step 4.2, those two $L^{\mathrm{\infty }}$ classes are distinct, so uniqueness fails. Therefore the $p=1$ representation theorem is false on this measure space.