1

I'm working through some literature from Geometric Measure Theory for an assignment paper and have got stuck at a step in the proof of the Federer-Volpert theorem. The reasoning is as follows: according to the BV coarea formula, the level sets $\{u>t\}$ have finite perimeter for $\lambda$-a.e. $t \in \mathbb{R}$. From this, the author concludes that we can find a dense subset $D$ of $\{ t \in \mathbb{R} \ \vert \ \{u > t \} \ \text{has finite perimeter} \}$. I'm guessing this must follow from some well-known fact in measure theory, but I can't put my finger on it. Any help would be greatly appreciated.

Othman El Hammouchi
  • 734
  • 4
  • 12
  • 2
    Well-known fact in topological measure theory: let $(X,d)$ be a complete separable metric space, and $\mu$ be a $\sigma$-finite measure on $X$ will full support. Then, every set of full measure is dense in $X$. See e.g. here. – AlephBeth Mar 20 '22 at 13:59
  • Thank you so much for the quick response! I wasn't familiar with this result. – Othman El Hammouchi Mar 20 '22 at 14:01
  • @AlephBeth : could you please provide a reference (possibly a book) where your statement explicitly appears? I agree that it is the consequence of the theory explained in every book on measure theory, but I have to put a reference in a paper and I do not find an explicit sentence! – Doriano Brogioli Oct 13 '22 at 07:57

1 Answers1

1

If $P(t)$ is any property that holds for almost every $t\in\mathbb R$, then $\{t\in\mathbb R:P(t)\}$ is dense in $\mathbb R$.

Proof. By assumption, $\lambda(\{t \in\mathbb R : \text{not}\ P(t)\}) = 0$, and hence $\{t\in\mathbb R : \text{not}\ P(t)\}$ contains no nontrivial interval $(a,b)$. Therefore, $\{t\in\mathbb R:P(t)\}$ intersects every nontrivial interval $(a,b)$, so it is dense. $\square$

Alex Ortiz
  • 24,844