When is $L^1 = (L^\infty)^\ast$?

I found this exercise in Cohn's Measure Theory:

Let $(X, \mathscr A, \mu)$ be a finite measure space. Show that the conditions

1. the map $T: L^1(X, \mathscr A, \mu) \to (L^\infty(X, \mathscr A, \mu))^\ast$ given by $g\mapsto T_g(f) = \int fg \, d\mu$ is surjective
2. $L^1(X, \mathscr A, \mu)$ is finite-dimensional
3. $L^\infty(X, \mathscr A, \mu)$ is finite-dimensional
4. there is a finite $\sigma$-algebra $\mathscr A_0$ on $X$ such that $\mathscr A_0\subset \mathscr A$ and such that each set in $\mathscr A$ differs from a set in $\mathscr A_0$ by a $\mu$-null set

are equivalent.

I figured out a way to show $2. \implies 4. \implies 3. \implies 2.$ and how these three conditions imply $1.$ What I'm having trouble with is how to get from 1. to either of the other three. If someone could provide a hint, I'd be grateful.

Thank you.

Solution 1:

Thanks to t.b.'s hint in the comments, I think I can prove $1. \implies 4.$ now.

Suppose 4. is not true for $(X, \mathscr A, \mu)$. Then there exists a sequence $B_n$ in $\mathscr A$ of pairwise disjoint sets with positive measure. Now choose a point $b_n \in B_n$ for each $n$ and let

$$C = \{f\in L^\infty(X, \mathscr A, \mu) \mid f \text{ is constant on $B_n$ for all $ n$ and } \lim_{n\to\infty} f(b_n) \text{ exists}\}$$

Let $\Lambda_0$ be the linear functional on $C$ given by $$\Lambda_0(f) = \lim_{n\to \infty} f(b_n)$$ Then $\Lambda_0$ is continuous on $C$ (in fact $\Vert \Lambda_0 \Vert = 1$) and we can extend $\Lambda_0$ to a linear functional $\Lambda$ on all of $L^\infty(X, \mathscr A, \mu)$ by an application of Hahn-Banach.

But this $\Lambda$ cannot be of the form

$$\Lambda(f) = \int_X fg \, d\mu$$

for any $g\in L^1(X, \mathscr A, \mu)$. Suppose there was such $g$. Let $A_n = \bigcup_{k=1}^n B_k$ and $A = \bigcup_{n=1}^\infty B_n$. Then we would have $$\int_X \chi_{A_n} g \, d\mu = \Lambda\left(\chi_{A_n}\right) = \lim_{m\to\infty} \chi_{A_n}(b_m) = 0$$ for all $n$. And therefore, by the monotone convergence theorem applied to $\chi_{A_n} g^+ \uparrow \chi_A g^+$ and $\chi_{A_n} g^- \uparrow \chi_A g^-$, we obtain

\begin{align} \int_X \chi_A g\, d\mu &= \int_X \chi_{A} g^+\, d\mu - \int_X \chi_{A} g^-\, d\mu \\\ &= \lim_{n\to\infty} \int_X \chi_{A_n} g^+\, d\mu - \lim_{n\to\infty} \int_X \chi_{A_n} g^- \, d\mu \\\ &= \lim_{n\to\infty} \int_X \chi_{A_n} g\, d\mu \\ &= 0 \end{align}

But on the other hand we have $\Lambda(\chi_A) = \lim_{n\to\infty} \chi_A(b_n) = 1$, a contradiction.

So this $\Lambda \in (L^\infty(X, \mathscr A, \mu))^\ast$ is not in the image of $T: L^1(X, \mathscr A, \mu)\to (L^\infty(X, \mathscr A, \mu))^\ast$.