If $E\subset \mathbb{R}$ is a set Lebesgue Measurable and $f:E \rightarrow \mathbb{R}$ a monotone function, show that $f$ is measurable.
I'm trying for hours with no progress.
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$\begingroup$ And what have you done so far? $\endgroup$– 5xumFeb 3, 2014 at 15:07
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$\begingroup$ What form does $f^{-1}((c,\infty))$ have, for $c\in \mathbb{R}$? $\endgroup$– Daniel FischerFeb 3, 2014 at 15:07
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$\begingroup$ @5xum Apply the definition of a function mensurable (in $\mathbb{R}$): $f$ is mensurable if $f^{-1}(I)$ is mensurable SET for any interval $I$ $\endgroup$– FelipeFeb 3, 2014 at 15:09
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$\begingroup$ OK. Try examining what $f^{-1}(I)$ can be, using the fact that $f$ can only have a countable set of points where it is not continous. $\endgroup$– 5xumFeb 3, 2014 at 15:10
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1$\begingroup$ I do not remember where I first found this fact, but it is stated on the wikipedia article on monotone functions: en.wikipedia.org/wiki/Monotonic_function The theorem itself is found here: en.wikipedia.org/wiki/Froda's_theorem $\endgroup$– 5xumFeb 3, 2014 at 15:42
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1 Answer
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Every monotone function is even Borel measurable, and in particular Lebesgue measurable. To see this, first let $f$ be an increasing function on $\mathbb{R}$. Let $a \in \mathbb{R}$.
Define $F_a=\{ x\ \in \mathbb{R} : f(x) \leq a \}$, and we want to show $F_a$ is a measurable set, from which we will conclude f is measurable.
Now if $F_a=\varnothing $ then obviously $F_a$ is measurable, and we are done.
So assume $F_a\not =\varnothing $, and let $x_0=\sup F_a$, There are 3 different options:
1)if $x_0=\infty$, then $F_a=\mathbb{R}$ is measurable.
2)if $x_0 \in F_a$ then $F_a=(-\infty,x_0]$.
3) if $x_0 \not \in F_a$ then $F_a=(-\infty,x_0)$.
and in any case $F_a$ is a borel set, and so f is measurable.
Now for f decreasing you may note that $-f$ is an increasing function, so it is measurable, and therefore $f$ is measurable too.
