0

I'm referring to Lebesgue measure here. If it helps, we can work in $[0,1]^n$ instead, since if a set has measure zero in each cube, then it has measure zero overall.

For example, I'm thinking of the set of invertible matrices (as a subset of $\mathbb R^{n^2}$), or the set of matrices with $n$ distinct eigenvalues. I'm pretty sure both of those are dense open sets. Is that sufficient to prove that the complement has measure zero?

It seems like it should be, but I can't seem to prove it, and now I'm worried it might not be true. I can't find a counterexample either.

If this is not a sufficient condition, I'd be interested to know if there is some extra fact that's true of the set of invertible matrices, that would make it sufficient.

Alex Eustis
  • 345
  • 1
    False in general (even if $n=1$), but true for real semialgebraic subsets such as invertible matrices. – Moishe Kohan Mar 26 '25 at 03:15
  • Thanks. How do you prove the second part? Is it a dimension argument, where $\det A =0$ defines a polynomial equation of homogeneous degree n, and so it's locally a manifold (submanifold?) of dimension $n^2-n$. Something like that? – Alex Eustis Mar 26 '25 at 03:36
  • 2
    This question was asked numerous times on MSE, take a look at this question and 6 questions linked to it. (But there are other duplicates as well.) – Moishe Kohan Mar 26 '25 at 04:13

1 Answers1

3

Consier an enumeration of the rationals of $(0,1)$, i.e. write $\mathbb{Q} \cap (0, 1) = \{q_1, q_2, \dots\}$. For each $q_i$ consider the open interval centered at $q_i$ with length $1/2^{i+1}$. The union of all such intervals is an open dense subset of $[0,1]$ of measure at most $1/2$. You can also generalize this to higher dimensions using a similar argument.

George Giapitzakis
  • 7,097