How can I show if an incremental function is Lebesgue measurable?

Question

I am trying to show that $f:\, I \to \mathbb{R}$ is Lebesgue measurable; where $I$ is an interval in the real axis and $f:\, I \to \mathbb{R}$ is an increasing function, but sadly I have no idea how to do this. How can I show this?

Answer 1

Hint

Prove that for $\alpha \in \mathbb R$,

$$\{x \in \mathbb R \mid f(x) < \alpha\}$$ is an interval, hence is measureable.

Answer 2

A function $f: I \to \mathbb{R}$ is measurable if $f^{-1}((a,b))$ is measurable for every $a,b \in [+\infty, -\infty]$ (Pardon this notation). You have a few cases here, either:

Lets say $I=[m,n]$

1. $f(I)\subset (a,b)$ then $f^{-1}((a,b))= I$

2. $\exists x \in I: f(x) < a$ and $f(m)< b$, then $f^{-1}((a,b)) = (l,m)$ for some $l \in I$, or $f^{-1}((a,b))$ is a point (call this case X). Because it is increasing

3. $\exists x \in I: f(x) > b$ and $f(n)> a$, analog to 2.

4. $(a,b) \subset [f(n),f(m)]$, then $f^{-1}((a,b))= (f(l),f(j))$ for some $j,l \in I$ or case X.

This is all because $f$ is an increasing function, so for any $x \in (a,b)$ you either have that $x$ has no pre-image in $I$ or that $\exists y \in I: f(y)=x$, then for every $z < x$ you have that $f^{-1}(z) < y$. So if you have $m,n$ such that $f(m), f(n) \in (a,b)$ then $f((m,n))$ is in $(a,b)$.