Re: do zero-measure sets map to zero-measure sets?

Thanks for the replies.

The problem I posted originally is actually much more general
than the specific problem I wish to solve. I should have given
my specific problem instead of this more general problem:

I have a map that takes a full-rank matrix of order m by n (with m<n)
to its kernel. The kernel is a linear subspace of dimension n-m in
R^n. If we identify each kernel as a point in the range, then the
range is a grassman manifold, which I will denote by G(n,n-m).
The grassman manifold G(n,n-m) is compact and differentiable, and
has dimension m*(n-m). It is equipped with a rotation invariant
Riemannian metric, which induces a distance function.
Hence we can compute Hausdorff measure.

Let S be a subset of G(n,n-m) with zero m*(n-m)-dimensional Hausdorff
measure. (To be more precise, it has sigma-finite m*(n-m)-1 dimensional
Hausdorff measure.)
I know that the preimage of each point in S has Lebesgue measure zero.
(In fact the preimage of each point in S has dimension m^2, which
is less than the dimension m*n of the domain.)

My question is: does the preimage of S have zero Lebesgue measure?

Thanks.

Simon