A few days ago, professor Wilkinson asked me the following question on google talk (while I was in Toronto):

“*Say that a set in is a -zero set for some integer if for every -dimensional subspace , saturating the set by planes parallel to yields a set of -dimensional Lebesgue measure zero.*

*How big can a -zero set be*?”

On the spot my guess was that the Hausdorff dimension of the set is at most . In deed this is the case:

First let’s note that -dimensional Lebesgue measure of the -saturated set is iff the dimensional Lebesgue measure of the projection of our set to the subspace orthogonal to is .

Hence the question can be reformulated as: If a set has all dimensional projection being zero sets, how big can the set be?

Looking this up in the book *‘The Geometry of Fractal Sets’* by Falconer, indeed it’s a theorem:

**Theorem:** Let compact, (Hausdorff dimension), let be the Garssmann manifold consisting of all -dimensional subspaces of , then

a) If , for almost all

b) If , has positive -dimensional Lebesgue measure for almost all .

In our case, we have some set with all -dimensional projection having measure , hence the set definitely does not satisfy b), i.e. it has dimensional at most . Furthermore, a) also gives that if we have a uniform bound on the dimension of almost all projections, this is also a bound on the dimension of our original set.

This is strict as we can easily find sets that’s dimensional and have all such projections measure . For example, take an subspace and take a full-dimension measure Cantor set on the subspace, the set will have all projections having measure .

Also, since the Hausdorff dimension of any projection can’t exceed the Hausdorff dimension of the original set, a set with one projection having positive measure implies the dimension of the original set is .

**Question 2**: If one saturate a -zero set by any smooth foliations with -dimensional leaves, do we still get a set of Lebesgue measure ?

We answer the question in the affective.

Given foliation of and -zero set . For any point , there exists a small neighborhood in which the foliation is diffeomorphic to the subspace foliation of the Euclidean space. i.e. there exists from a neighborhood of to where the leaves of are sent to , .

By restricting to a small neighborhood (for example, by taking to be half of the origional ), we may assume that is bi-Lipschitz. Hence the measure of the -saturated set inside of is the same as saturated by parallel -subspaces inside . Dimension of is the same as dimension of which is , if the inequality is strict, then all projections of onto dimensional subspaces has measure i.e. the saturated set by -planes has dimensional measure .

…to be continued