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?

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    $\begingroup$ 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. $\endgroup$ – David C. Ullrich Jul 6 '15 at 17:01
  • $\begingroup$ @DavidC.Ullrich: an interval like $(c, +\infty)$ or $[c, +\infty)$? $\endgroup$ – Bear and bunny Jul 6 '15 at 17:07
  • $\begingroup$ See here: math.stackexchange.com/questions/229497/… $\endgroup$ – Math1000 Jul 6 '15 at 17:13
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    $\begingroup$ Yup${}{}{}{}{}{}$ $\endgroup$ – David C. Ullrich Jul 6 '15 at 17:14
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    $\begingroup$ @Math1000: Awesome. A more general idea. Thanks. $\endgroup$ – 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.

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