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

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?

• It is sufficient, but I have no appropriate reference at hand. Every bounded linear functional on $L_\infty(\mu)$ can be represented as integration with respect to a finite finitely additive signed measure absolutely continuous with respect to $\mu$. Each such finitely additive signed measure has a Jordan decomposition and the rest is a standard characterization of countable additivity for bounded finitely additive set functions by the measure of sequences of sets decreasing to the empty set decreasing to zero. May 27 '18 at 19:53
• (But why "hypothesis"? You are stating it as a conjecture, aren't you?) May 28 '18 at 3:46
• @PietroMajer, actually I don't feel the difference, my English is not that good. May 28 '18 at 6:25
• My apologies, I didn't mean to be nitpicking :) May 28 '18 at 7:20
• Pietro, it's OK, I did not suspect this. I read about the difference here: mathforum.org/library/drmath/view/52249.html So now I know about it. :) May 28 '18 at 7:41

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.

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 $\mathcal{M}_* = L^1(X,\Sigma,\mu)$.

• These non $\sigma$-finite examples are also counterexamples to $L^\infty(X)$ being the dual of $L^1(X)$, aren't they? May 28 '18 at 7:46
• @PietroMajer The first one is, the second one isn't (which is why I inncluded a second one). $L^\infty(X)$ is the dual of $L^1(X)$ iff $X$ is localizable as a measure space. See, for instance: jstor.org/stable/2372178 May 28 '18 at 10:58
• @PietroMajer It's also relatively easy to prove that $\ell^\infty(X)$ is the dual of $\ell^1(X)$ for all sets $X$, once you know it's true, even without the more general theorem. May 28 '18 at 11:05
• Robert, I don't understand something in your second example. This measure $\mu:{\mathcal P}(X)\to [0,1]$, it must be finite, isn't it? May 31 '18 at 15:33
• @SergeiAkbarov Quite right, I'll just fix it. May 31 '18 at 18:12

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.

• billabong, could you, please, give some references? May 27 '18 at 20:25
• simply google under Saks spaces. It even has an MOS classification number. The paper of Saks is also easy to trace. May 27 '18 at 20:29
• The article by Saks is simply called "On some functionals". It is so celebrated that you will get an immmediate hit as a pdf file by simply googling the author's name and the title. It appeared in one of the AMS journals in the 30's of the last century. May 27 '18 at 20:35
• Billabong, excuse me I could not reply earlier. That is interesting, but I did not understand, what the result is exactly. If $\tau$ is a topology on $L_\infty(\mu)$ such that $(L_\infty(\mu),||\cdot||_\infty,\tau)$ is a Saks space, then $L_1(\mu)$ is the space of functionals on $L_\infty(\mu)$ continuous with respect to $\tau$? Is that what you mean? And where is this written? May 28 '18 at 6:36
• If what I guess is what you had in mind (or if there is another similar criterion), then you should roll back your answer and formulate the result explicitly. You can also contact me directly by email, if you have doubts. May 28 '18 at 6:53