# The complement of $L_1(0,1)$ in $L_1(0,1)^{**}$

Let $$\mu$$ be a finite measure, like the Lebesgue measure in $$(0,1)$$. It is well-known that $$L_1(\mu)$$ and its second dual $$L_1(\mu)^{**}$$ are Banach lattices, $$L_1(\mu)$$ is a projection band in $$L_1(\mu)^{**}$$, and we can write $$L_1(\mu)^{**}=L_1(\mu)\oplus L_1(\mu)^d$$, where $$L_1(\mu)^d = \{x^{**}\in L_1(\mu)^{**} : |x^{**}|\wedge |x|=0 \textrm{ for each }x\in L_1(\mu)\}$$.

In "Linear Operators, Part I" 1988 by Dunford and Schwarz (IV.8.16 on page 296), there is a description of $$L_1(\mu)^{**}$$ as 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.

I am interested in a description of the finitely additive measures that correspond with elements of $$L_1(\mu)^d$$.

• Could you clarify what you mean by "a description" in your last sentence? I'm not sure whether one can expect anything more explicit than what you cite from Dunford and Schwarz. Commented Sep 20, 2022 at 18:06
• I am looking for a condition that characterizes the elements of $L_1(\mu)^d$, but I am not sure what kind of condition I could expect. Commented Sep 20, 2022 at 18:28
• The idea is to decompose a f.a. measure $\mu$ as $\mu = \mu_1+\mu_2$ where $\mu_1$ is countably additive and $\mu_2$ is [in the sense you are looking for] as far from countably additive as possible. Called "purely finitely additive" in the literature. Yoshida & Hewitt, 1952. Commented Sep 20, 2022 at 18:28
• In principle, $L_1(\mu)$ admits many complements in $L_1(\mu)^{**}$. I am interested in one which is a Banach lattice with the order induced by $L_1(\mu)^{**}$. I would have to check the Yoshida-Hewitt decomposition. Commented Sep 20, 2022 at 18:57
• Look at Theorem I.5.8 in Diestel-Uhl. It is the Yosida-Hewitt theorem in the vector valued setting. Commented Sep 20, 2022 at 19:36

If I understand your question correctly, you are interested in a convenient description of $$L_1^d(\mu)$$. The finite additive measures are very inconvenient to work with. But Gelfand's representation allows to avoid them. According to

Kôsaku Yosida and Edwin Hewitt, Finitely additive measures, Trans. Amer. Math. Soc. 72 (1952), 46-66 (see theorem 4.3)


there exists a compact Hausdorff topological space $$X$$ such that Banach spaces $$L_\infty(\mu)$$ and $$C(X)$$ are isomorphic, $$L_\infty(\mu)\simeq C(X)$$. Therefore, $$L_\infty^*(\mu)\simeq C^*(X)$$. The last one is the space of Radon measures (regular countable additive Borel measures) on $$X$$ by the Riesz-Markov-Kakutani representation theorem, $$L_\infty^*(\mu)\simeq\mathrm{rca}(X)$$.

https://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space

So instead of nasty finite additive measures in $$L_\infty^*(\mu)$$ we have nice countable additive measures on a compact Hausdorff space $$X$$. So $$L_1(\mu)$$ is identified with measures on $$X$$ that are absolutely continuous w.r.t. $$1\in C(X)$$, and $$L_1^d(\mu)$$ are whose that are singular to $$1\in C(X)$$.

The space $$X$$ and isomorphism $$L_\infty(\mu)\simeq C(X)$$ can be described more or less explicitly. Namely, $$X$$ is the space of characters $$\chi$$ on $$L_\infty(\mu)$$, i.e. $$\chi\in L_\infty^*(\mu)$$ and $$\chi(fg)=\chi(f)\chi(g)$$ for any $$f,g\in L_\infty(\mu)$$. The topology on $$X$$ is the restriction of the weak$$^*$$ topology on $$L_\infty^*(\mu)$$. The isomorphism $$L_\infty(\mu)\simeq C(X)$$ is given by

$$f\in L_\infty(\mu)\quad\mapsto\quad F\in C(X)$$

where $$F(\chi) = \chi(f)$$.