A characterization of $L_1(\mu)$ in $L_\infty(\mu)^*$

Question

Let $\mu$ be a finite positive measure on a set $M$: $$ \mu(M)<\infty. $$ As is known, the Banach dual space $L_\infty(\mu)^*$ to the space $L_\infty(\mu)$ contains $L_1(\mu)$, but (excluding some trivial situations) does not coincide with $L_1(\mu)$: $$ L_1(\mu)\subseteq L_\infty(\mu)^*, \qquad L_1(\mu)\ne L_\infty(\mu)^*. $$

Is there a reasonable description of $L_1(\mu)$ as a space of functionals on $L_\infty(\mu)$?

For example, is the following true?

Conjecture. A functional $f\in L_\infty(\mu)^*$ belongs to $L_1(\mu)$ if and only if $$ f(x_n)\underset{n\to\infty}{\longrightarrow} 0 $$ for each sequence of functions $x_n\in L_\infty(\mu)$ which is bounded in $L_\infty(\mu)$ as in a Banach space $$ ||x_n||_\infty\le 1, $$ and converges to zero almost everywhere $$ x_n\overset{\text{almost everywhere}}{\underset{n\to\infty}{\longrightarrow}}0. $$

This condition is necessary due to the Lebesgue's dominated convergence theorem. Is it sufficient?

Answer 1

It is shown in "Linear Operators, Part I" 1988 by Dunford and Schwarz as IV.8.16 on page 296 that the dual of $L_\infty(\mu)$ can be identified with the space of finitely additive bounded signed measures that are absolutely continuous with respect to $\mu$ (with the variation norm).

Every such finitely additive measure that is countably additive corresponds to an element of $L_1(\mu)$ by the Radon-Nikodym theorem. So the problem reduces to identifying countable additivity. If $\langle A_n\rangle$ is a disjoint sequence of measurable sets, we have that indicator functions of $\bigcup_{n=1}^\infty A_n\setminus\bigcup_{n=1}^m A_n$ converge pointwise to zero, so countable additivity follows from the given condition.

Answer 2

The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption.

However, the suggested criterion is not (provably) sufficient in general. I can give two examples. The first is quite pathological, and the second is not, but goes beyond ZFC.

(1) Let $X$ be an uncountable set, and take $\Sigma$ to be the countable-cocountable $\sigma$-algebra (the one generated by singletons). Take $\kappa$ to be the counting measure, i.e. a set $S \in \Sigma$ has measure $|S|$ if it is finite and $\infty$ if it is of any infinite cardinality. A function is integrable with respect to this measure iff it is an element of $\ell^1(X)$ (the small $\ell$ is intentional). Therefore a functional in $L^\infty(X,\Sigma,\kappa)$ that is in the image of $L^1(X,\Sigma,\kappa)$ must take a nonzero value on some $\chi_{\{x\}}$ for some $x \in X$.

Now consider the following measure $\nu : \Sigma \rightarrow [0,1]$. It takes the value $0$ on all countable sets and $1$ on all cocountable sets. A quick check shows that it is countably additive, and it is absolutely continuous to $\kappa$. But for all $x \in X$, $\int_X \chi_{\{x\}} \mathrm{d}\nu = 0$, so integration against it does not come from an element of $L^1(X,\Sigma,\kappa)$, even though it satisfies your criterion by the dominated convergence theorem.

(2) The reason why that space is somewhat pathological is that it is not localizable, i.e. $L^\infty(X,\Sigma,\kappa)$ is not a W$^*$-algebra (isomorphic to a von Neumann algebra). The following is a way of getting a localizable counterexample. Suppose that there exists a real-valued measurable cardinal, i.e. that there exists a set $X$ and a countably additive probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons. Such a measure is necessarily not completely additive, because it would then be zero (and the set $X$ is necessarily uncountable for the same reason). So there exists a family of sets $(S_i)_{i \in I}$ where $\mu(S_i) = 0$ for all $i \in I$, but $\bigcup_{i \in I}S_i = X$ (for instance, take the family of all singletons).

Now recall the following facts. The spaces $\ell^p(X)$ are exactly the spaces $L^p(X,\mathcal{P}(X),\kappa)$, where $\kappa$ is the counting measure (on $\mathcal{P}(X)$ this time), and each element of $\ell^1(X)$ defines a completely additive signed measure $\nu : \mathcal{P}(X) \rightarrow \mathbb{R}$ by integration (because all $\phi \in \ell^1(X)$ have countable support). By the dominated convergence theorem, the element of $L^\infty(X,\mathcal{P},\kappa)^*$ defined by $f \mapsto \int_X f \mathrm{d}\mu$ satisfies your criterion, but cannot be equal to any $f \mapsto \int_X f\phi\mathrm{d}\kappa$ for any $\phi \in \ell^1(X)$ because $$ \lim_{i \to \infty} \int_X \chi_{S_i} \mathrm{d}\mu = \lim_{i \to \infty} 0 = 0 $$ for all $i \in I$, but for all $\phi \in \ell^1(X)$ that are density functions of probability measures: $$ \lim_{i \to \infty} \int_X \chi_{S_i} \phi \mathrm{d}\kappa = \int_X \chi_X \phi \mathrm{d}\kappa = 1 $$

Conclusion

If you wish to go beyond the $\sigma$-finite case to the localizable case, your criterion is not far off. Essentially all you need to do is replace sequences by nets. The image of $L^1(X,\Sigma,\mu)$ in $L^\infty(X, \Sigma,\mu)^*$ is the normal linear functionals, the span of those positive linear functionals that preserve suprema of bounded nets. See Sakai's C$^*$-algebras and W$^*$-algebras, Theorem 1.13.2, taking $\mathcal{M} = L^\infty(X,\Sigma,\mu)$ and $\mathcal{M}_* = L^1(X,\Sigma,\mu)$.

Answer 3

The space $L^\infty$, provided with its intrinsic norm and the $L^1$ norm, is an example of a Saks space (two normed space) for which there is an extensive theory. If one provides it with the so-called strict or mixed topology, i.e., the finest locally convex topology which agrees with that of $L^1$ on the unit ball, then its dual conists of the linear forms which are $L^1$-continuous on the unit ball. In your situation this is precisely the closure of $L^\infty$ (as the dual of $L^1$) in the norm dual of $L^\infty$ and and so you get $L^1$ as required. This follows from the general theory of Saks spaces exposed in the reference in my comment below.