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:


1. 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.
2. 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.



Which sigma-ideals in a sigma-algebra are ideals of null sets?
First of all, one should mention that not every triple (X,B,μ) (i.e., what is often called a *measure space*)
satisfies the property that its C\*-algebra of bounded functions is a von Neumann algebra (= W\*-algebra) or that the map L^∞→(L\_1)\* is an isomorphism.
One has to impose additional conditions to ensure that this is true.


If we do not assume such a condition, Joseph Van Name's answer solves the problem.


However, if one is interested in a condition that would guarantee
that (X,B,I) (the σ-ideal I is often known as a *measure class*) comes from a triple (X,B,μ) whose algebra of bounded
functions is a von Neumann algebra (and pretty much all of measure theory happens in this restricted setting because otherwise almost every nontrivial theorem would fail), then
an answer was given in 1951 by Irving Segal in his paper
“Equivalences of measure spaces”.


Segal proved that several seemingly unrelated properties are equivalent for a triple (X,B,I) (i.e., a measure space equipped with a measure class).
Two of these properties have already been named: L^∞(X,B,I) is a von Neumann algebra and the canonical map L^∞(X,B,I)→L^1(X,B,I)\* is an isomorphism (i.e., the Riesz representation theorem is satisfied).
Among other equivalent properties one can find the Radon-Nikodym theorem,
the Hahn decomposition theorem, and the fact that the Boolean algebra B/I is complete.
The full list and additional details can be found in the answer <https://mathoverflow.net/a/20820/402>.


Such triples (X,B,I) are known as localizable measurable spaces
and they can be organized into a category whose morphisms
(X,B,I)→(Y,C,J) are equivalence classes of measurable maps X→Y such that
the preimages of elements of J belong to I (this true in the complete case,
in the general case a slightly more complicated definition is necessary,
as described in the link).


This category turns out to be contravariantly equivalent to the category
of commutative von Neumann algebras (i.e., a version of Gelfand duality is true for von Neumann algebras), which further stresses the
importance of the localizability condition.
More information about this category and further links can be found in this answer:
<https://mathoverflow.net/a/49542/402>.


