1

Suppose that $\varphi : [a, b] \to \mathbb{R}$ is a simple function and let $\varepsilon > 0$ be given. Prove that there is a step function $g : [a, b] \to \mathbb{R}$ such that $g(x) = \varphi(x)$ except on a set of measure $\varepsilon$.

I was wondering if I could get a hint.

frito
  • 133
  • A Lebesgue measurable set can be approximated by an open set. An open set can be approximated by a finite union of open intervals (an open set is a countable union of intervals). – David Mitra Apr 25 '13 at 20:54
  • @DavidMitra: your last point, the approximation: what does it follow from? – Alex Apr 25 '13 at 21:42
  • @Alex An open set $O$ of $\Bbb R$ can be written as a countable union of disjoint open intervals (see this). If $O$ has finite measure, then finitely many of these will give most of the mass of $O$. – David Mitra Apr 25 '13 at 22:18

0 Answers0