Chains of null sets

Noam Elkies made a cute observation on his site, which is that there are chains of null subsets of [0, 1] whose union is not null. Proof: Take a maximal chain. The union of this can't be null or it would contradict the maximality of the chain.

He then went on to ask how 'large' their union can be: Under the continuum hypothesis it can be all of [0, 1]. Do you need the continuum hypothesis for this? Without it how large can the union be?

I tinkered around with applying some cardinal invariants to the problem and came up with a nearly complete solution. I'm going to be posting a pdf of it at some point, but here are the highlights:

The solution is for all intents and purposes purely combinatorial. I used essentially no measure theory in proving it except for one lemma which has a comparably easy category analogue.

It basically depends on three cardinal invariants. add, cov and non. The additivity is the smallest cardinal k such that there are k many null sets whose union is not null. cov is the smallest cardinality of a covering of [0, 1] by null sets. non is the smallest cardinality of a non-null set.

We have aleph_1 <= add <= non, cov <= 2^aleph_0. add is regular, the other two don't have to be, and no inequalities between non and cov are provable in ZFC. (I screwed up and claimed in my email to Elkies that in fact non <= cov and they were both regular. This is false, but it only broke a minor part of my conclusions).

It's relatively easy to see that if non = 2^aleph_0 or add = cov then you can get a chain whose union is [0, 1]. What's slightly harder is that if such a chain exists then you can find some regular cardinal k (the cofinality of the chain) with cov <= k <= non. So, because you can have non < cov, you can't always find such a chain.

So, in the absence of one with union [0, 1], how big can it be? Can it be measurable? (If it is measurable then of course it has measure > 0). Turns out not, and this is where the tiny bit of measure theory comes in. If A is measurable and m(A) > 0 then mu(A + Q) = 1 (addition is mod 1). So, if we have our chain L_t of null sets whose union is measurable and of positive measure, then replacing each L_t with L_t + Q gives a chain of null sets whose union is of full measure. Then adding in the complement we get it to be all of [0, 1].

The only question I have remaining which I'm not sure about is whether or not the existence of such a regular cardinal is equivalent to the existence of such a chain. I doubt it, but I'm not really sure and probably lack the knowledge of forcing needed to understand a proof either way.