How can I show that the following function is measurable

I just need a few solid hints.

The problem:

Suppose $I$ is an interval and define an increasing function on $I$ by $f: I \rightarrow \mathbb{R}$. Show that $f$ is a measurable function by first showing that for each natural number $n$, the strictly increasing function $x \rightarrow f(x) + \frac{x}{n}$ is measurable and then taking pointwise limits.

Here's what I have so far. Assuming that $f(x) + \frac{x}{n}$ is measurable, for each $n$, we have a sequence of measurable functions on our hands. By a previous problem, I know that the $inf$ of a sequence of measurable functions is measurable. So if I take a pointwise limit, by fixing $x$ and letting $n$ tend to infinity, I will get the $inf$ of the sequence, which is $f(x)$ and thus $f(x)$ is measurable.

My problem is, how do I first show that $f(x) + \frac{x}{n}$ is measurable for each $n$?

Thank you.

• Can I say that since the function must necessarily be invertible, the inverse image of an interval (open set) under $f$ is measurable in $I$?, thus making the function measurable?? Oct 27, 2016 at 22:58
• you shouldn't take the infimum, as $x$ might be negative. Just take the limit (or liminf or limsup). Oct 27, 2016 at 23:00
• Okay, but I don't see how limsup would be appropriate here. The functions are approaching $f(x)$ from the top. Oct 27, 2016 at 23:04
• $f_n(x) = f(x) + x/n$ converges pointwise to $f(x)$, so the limit equals the limit inferior equals the limit superior. Oct 27, 2016 at 23:05
• Possible duplicate of Are Monotone functions Borel Measurable? Oct 27, 2016 at 23:34

You can show instead that for every $r$, $f^{-1}(-\infty,r)$ is an interval, unbounded to the left (so either of the form $(-\infty,a]$ or $(-\infty,a)$, with $a=\sup f^{-1}(-\infty,r)$): Indeed, if $x\in f^{-1}(-\infty,r)$ and $y<x$ then $f(y)\leq f(x)<r$, so $y\in f^{-1}(-\infty,r)$.
I don't think that using $f_n(x)=f(x)+x/n$, which is strictly increasing, helps in any way, because $f_n^{-1}(-\infty,r)$ and $f_n^{-1}(-\infty,r]$ can be either of the form $(-\infty,a)$ or $(-\infty,a]$ (e.g., take $f(x)=\operatorname{sign}(x)$ for $x\neq 0$, define $f(0)=0$ and look at preimages $f_n^{-1}(-\infty,0]$ and $f_n^{-1}(-\infty,0)$. Do the same by defining $f(0)=1$ or $f(0)=-1$, and the preimages will have different forms.)