Show that if $E\subset\mathbb{R}$ is a measurable set, so $f:E\rightarrow \mathbb{R}$ is a measurable function.

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.

• And what have you done so far? – 5xum Feb 3 '14 at 15:07
• What form does $f^{-1}((c,\infty))$ have, for $c\in \mathbb{R}$? – Daniel Fischer Feb 3 '14 at 15:07
• @5xum Apply the definition of a function mensurable (in $\mathbb{R}$): $f$ is mensurable if $f^{-1}(I)$ is mensurable SET for any interval $I$ – Felipe Feb 3 '14 at 15:09
• 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. – 5xum Feb 3 '14 at 15:10
• 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 – 5xum Feb 3 '14 at 15:42

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.