A question concerning Borel measurability and monotone functions I came across the following exercise in my self-study:
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is monotone, then $f$ is Borel measurable.
I am unsure about how to proceed from the hypothesis to give the requisite proof, in particular how sensitive I should be to proof by cases. Would anyone visiting have any suggestions, or be up for proving this interesting little result?
 A: A function $f:\mathbb R\to\mathbb R$ is Borel if and only if for all $a\in\mathbb R$, the set $\{x\in\mathbb R:f(x)>a\}$ is a Borel subset of $\mathbb R$.  
Suggestion: Think about what possible types of sets you can get for $\{x\in\mathbb R:f(x)>a\}$ when $f$ is a monotone function.  You may want to conjecture with the aid of examples before trying to prove your conjecture.  The sets you get should be easily confirmed to be Borel.
A: Proposition
Let $X$ be a Borel subset of $\mathbb{R}$.
Let $\mathfrak{B}$ the set of Borel subsets of $X$.
$(X, \mathfrak{B})$ is a measurable space.
Let $f\colon X \rightarrow [-\infty, \infty]$ be a monotone function.
Then $f$ is measurable.
Proof:
Without loss of generality, we can assume that $f$ is non-decreasing.
Let $a \in (-\infty, \infty)$.
Let $E_a = \{x \in X; f(x) > a\}$.
Let $c = \inf E_a$.
It suffices to prove that $E_a$ is a Borel subset. 
Case 1. $c \in E_a$.
Let $x \in X \cap [c, \infty)$
Since $c \le x$, $a < f(c) \le f(x)$.
Hence $x \in E_a$.
Conversely suppose $x \in E_a$.
Since $c \le x, x \in X \cap [c, \infty)$.
Therefore $E_a = X \cap [c, \infty)$.
Case 2. $c$ does not belong to $E_a$.
Let $x \in X \cap (c, \infty)$.
There exists $y \in E_a$ such that $c < y < x$.
Since $a < f(y) \le f(x), f(x) > a$.
Hence $x \in E_a$.
Conversely suppose $x \in E_a$.
Since $c \le x$ and $c$ does not belong to $E_a$, $x \in X \cap (c, \infty)$.
Therefore $E_a = X \cap (c, \infty)$.
QED
