Property holding almost everywhere implies existence of dense subset?

Question

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.

Answer 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. $\square$