Summary: The critical set goes to a set of measure zero.
Proposition. Suppose S is an r-dimensional linear subspace of Rn
and b Rn
. Then for any R > 0 and
any > 0 we have
Ln
({y Bb(R) : dist (y, b + S) }) 2n
Rr n-r
.
Proof. Translating by -b if necessary we may assume that b = 0. Rotating S if necessary we may assume
that S = Rr
× {0} where we identify Rn
with Rr
× Rn-r
and where we make use of the fact that
Ln
(L[A]) = |det L|Ln
(A) for any Lebesgue measurable subset of Rn
.
Finally, we observe that

Source: Allard, William K. - Department of Mathematics, Duke University

Collections: Mathematics