# Is a monotone function defined on any kind of interval measurable?

Definition of Measurable set: A set $$E$$ measurable if $$m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$$ for every subset of $$A$$ of $$\mathbb R$$.

Definition of Lebesgue measurable function: Given a function $$f: D \to \mathbb R$$, defined on some domain $$D$$, we say that $$f$$ is Lebesgue measurable if $$D$$ is measurable and if, for each real $$a$$, the set $$\{x\in D: f(x) \gt a\}$$ is measurable.

So suppose $$f(x)$$ defined on $$[a,b]$$ is a monotone function, then $$f(x)$$ is a measurable function because $$\{x∈[a,b] \mid f(x) > t, t ∈ \mathbb R\}$$ must be one of the three situations--interval(closed or half open half closed), a single point set or $$\emptyset$$ while each of them is a measurable set.

My question is how about monotone $$f(x)$$ defined on $$(a,b]$$ or $$(a,b)$$ or $$[a,b)$$? Do kinds of interval domains affect $$f$$'s Lebesgue measurable?

• If $E\subset\mathbb R$ is any measurable set and $f:E\to\mathbb R$ is monotone then $f$ is measurable. You can easily work this out - the set of points in $E$ where $f>a$ is the intersection of $E$ and a ____, so it's measurable because the intersection of two measurable sets is measurable. – David C. Ullrich Jul 6 '15 at 17:01
• @DavidC.Ullrich: an interval like $(c, +\infty)$ or $[c, +\infty)$? – Bear and bunny Jul 6 '15 at 17:07
• – Math1000 Jul 6 '15 at 17:13
• Yup${}{}{}{}{}{}$ – David C. Ullrich Jul 6 '15 at 17:14
• @Math1000: Awesome. A more general idea. Thanks. – Bear and bunny Jul 6 '15 at 17:17

Cite Math1000's comment:$f: \mathbf{R} \rightarrow \mathbf{R}$ monotone increasing $\Rightarrow$ $f$ is measurable.
This is a more general question and Cass's answer is pretty clear and concise. And I cite his here " If $f$ is increasing, the set {$x:f(x)>a$} is an interval for all $a$, hence measurable. By definition (Royden's), the function $f$ is measurable".
Combined with David C. Ullrich's comment, since $E$, be any of (a,b] or (a,b) or [a,b), measurable, $E ∩$ the interval is measurable that imply $f$ is Lebesgue measurable.