The intuition comes from the fact that every open set in $R^n$ must contain a basis for $R^n$. Now, I want to extend this to a set of positive measure. But I don't know how to start it.
Asked
Active
Viewed 52 times
2
Lin Xuelei
- 561
-
4Hint: Consider a counterexample and let $V$ be its span, which must then be a vector space of dimension $<n$. Derive a contradiction. – lulu Oct 29 '24 at 13:34
-
oh, right! any subspace of $R^n$ with dimension less than $n$ must be measure zero. Thanks – Lin Xuelei Oct 29 '24 at 13:37
1 Answers
2
Now I get it. Thanks for lulu's hint. I'll show it by contradiction. Let $S\subset R^n$ be a set of positive measure. Suppose $S$ does not contain a basis for $R^n$. Then, $S$ must lie in a subspace $V$ of $R^n$ and $\dim(V)<n$. Then, $m(S)\leq m(V)=0$ implies that $S$ is measure-zero. A contradiction.
Lin Xuelei
- 561