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

Question

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 $L_1(\mu)^d$.

Answer 1

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