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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.

Community
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Sam
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  • This is just an assumption. I don't have time to check this idea. $L^1$ - is separable, then does $\mathrm{Im}(T)=(L^\infty)^*$. Hence $L^\infty$ is separable. This is possible only if $L^\infty$ is finite dimensional. – Norbert May 29 '12 at 03:30
  • @Norbert: Thank you for the thought, but $L^1$ is not really separable for an arbitrary measure space $(X, \mathscr A, \mu)$, is it? I think your approach will work as long as $\mathscr A$ is countably generated, though. – Sam May 29 '12 at 03:57
  • If you toss away the axiom of choice there are some situations where $(L^\infty)^\ast=L^1$. See my answer here and the references there. – Asaf Karagila May 29 '12 at 06:17
  • I think the following should work: If 4. does not hold, then there is a countable collection ${B_n}_{n \in \mathbb{N}} \subset \mathcal{A}$ of pairwise disjoint sets of positive measure. Define $S\colon\ell^{1} \to L^{1}(\mu)$ by sending $e_n$ to the characteristic function of $B_n$. Show that $S$ is injective and that $S(\ell^1)$ is closed subspace of $L^1$. Thus, $L^1$ contains a non-reflexive subspace and hence $L^1$ is itself not reflexive. As $T$ is the canonical inclusion $L^1 \to (L^{1})^{\ast\ast}$, it is not surjective, hence 1. does not hold. – t.b. May 29 '12 at 08:04
  • @t.b.: Thank you for this idea! How do you construct the sequence ${B_n}_{n\in \mathbb N}$? I think this might be a crucial step I've been missing so far. I've only managed to get a sequence, such that $B_i \Delta B_j$ has positive measure for all $i\ne j$ (which is pretty much a restatement of the negation of 4.). – Sam May 29 '12 at 08:22
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    I haven't thought about it very deeply, but I'd start with throwing away the atoms: decompose $\mu$ into its atomic and atomless parts $\mu = \mu_a + \mu_c$. By the negation of 4 we either have countably many atoms (in which case we're done) or $\mu_c \neq 0$. But it is straightforward to decompose an atomless measure in the desired way: Given $B_{1},\dots,B_n$ pairwise disjoint and $\mu_c(B_i) \neq 0$ for all $i$ then we can split $B_n$ non-trivially into $B_{n}' \cup B_{n+1}'$ and proceed by induction. – t.b. May 29 '12 at 10:23
  • @t.b.: Great! Thank you once again. :) – Sam May 29 '12 at 12:36

1 Answers1

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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 $T: L^1(X, \mathscr A, \mu)\to (L^\infty(X, \mathscr A, \mu))^\ast$.

Sam
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  • This looks good. Another way to put your construction would be: we have an isometric embedding $\ell^{\infty} \to L^\infty(X,\mathcal{A},\mu)$ given by sending $(a_n){n \in \mathbb{N}}$ to $\sum{n \in\mathbb{N}} a_n \chi_{B_n}$. Thus, $L^\infty(X,\mathcal{A},\mu)$ is not reflexive and hence $L^1(X,\mathcal{A},\mu)$ can't be reflexive either. (Your construction amounts to extending a standard element of $(\ell^{\infty})^\ast \smallsetminus \ell^1$ to all of $L^\infty$ via this embedding). – t.b. May 29 '12 at 12:41
  • @t.b.: I guess I'm not too comfortable with functional analysis yet, so I wanted to go through the construction to make sure I know what I'm doing. But your way of looking at it is much more conceptual, of course. So thanks for pointing it out! – Sam May 29 '12 at 13:05
  • @Norbert: I tried just now, but it says that I need to wait another 21 hours. – Sam May 30 '12 at 06:00