My question is motivated, to be somewhat vague, by an attempt to see how much a measure space is defined by the set of null sets. In other words, assume we are not given a concrete measure on a space but are just told which sets are null - what can we say about the measure space? Note that, for example, membership in $L^\infty$ depends only on the questions of which sets are null.

Concretely, let $ \left (X, \mathcal{B} \right ) $ be a measurable space. A $\sigma$-ideal $\mathcal{I}$ is a subset of $\mathcal {B}$ which is closed under countable unions, contains the null set, and such that whenever $I \in \mathcal{I}$ and $B \in \mathcal{B}$, $I \cap B \in \mathcal{I}$. The standard example of $\sigma$-ideal is the family of null sets in a measure space $\left (X, \mathcal{B}, \mu \right) $.

Question: Given an arbitrary $\sigma$-ideal in a $\sigma$-algebra, when is it the family of null sets for some measure on the space?

I have not been able to find an answer to this question that does not amount to a tautology, for example, assuming that there exists a positive-definite function on $\mathcal{B}$ which essentially satisfies the axioms of a measure.

My approach to the question so far has been to attempt to define an analogue of $L^\infty$ for this case: for a measurable $f:X \to \mathbb{C}$, define the essential supremum $ \left \|f \right \|_\infty$ to be the infimum over all positive numbers $M$ such that $|f(x)| \le M$ outside of some set in $\mathcal{I}$, and let $L^\infty (X, \mathcal{B}, \mathcal{I})$ be the space of all functions with finite essential supremum.

It is easy to check that $L^\infty (X, \mathcal{B}, \mathcal{I})$ is always a Banach space, and by defining the $*$ operation to be conjugation as usual, it in fact becomes a $C^*$-algebra. If we knew that it is a Von-Neumann algebra (actually, a $W^*$-algebra, meaning it has a predual), it would be easy to conclude that $\mathcal{I}$ is the family of null sets for a measure, since then our space would be isomorphic to $L^\infty (Y, \mathcal{S}, \mu)$ and this isomorphism would necessarily take characteristic functions to characteristic functions.

From here, I can think of two ways to proceed:

Define also an analogue of $L^1$ in this case: since Radon-Nikodym allows us to identify $L^1$ with the space of all finite measures which are absolutely continuous wrt to a given measure, we define $L^1(X,\mathcal{B},\mathcal{I})$ to be the set of all measures which are zero on all sets in $\mathcal{I}$. The first problem with this is that in general, $(L^1)* \neq L^\infty$. This can be remedied by making a very reasonable assumption on $\mathcal{I}$, which I can say more about if needed, but even then, I don't see how to proceed.

Find a representation of $L^\infty(X,\mathcal{B},\mathcal{I})$ on a separable Hilbert space. However, I don't know how to construct a suitable Hilbert space without some sort of tautological assumption on the $\sigma$-algebra of the style I have written above.

Any reference or ideas would be greatly appreciated.