Fall 2017 - Problem 1

Solution.

Since $f$ is monotone, it is continuous except at countably many points $D$, hence Borel. For each $x \in D$, define $f_+\p{x} = \lim_{y \to x^+} f\p{y}$ and $f_-\p{x} = \lim_{y \to x^-} f\p{y}$. By modifying $f$ on a countable set, we may assume without loss of generality that $f_-\p{x} = f\p{x}$, i.e., $f$ is left continuous. Then given $x \in D$, let

where $U_x = \set{f_+\p{x}}$ if $f_+\p{x} \in f\p{\R}$ and $\emptyset$ otherwise.

We claim that $f\p{\R}^\comp = \bigcup_{x \in D} I_x$. It's clear that "$\supseteq$", since $f$ is monotone and $U_x$ handles the upper endpoint. Conversely, suppose $y \notin f\p{\R}$. Let $x_- = \sup\,\set{x \mid f\p{x} < y}$ and $x_+ = \inf\,\set{x \mid f\p{x} > y}$. We claim that $x_- = x_+$: otherwise, $x_- < x_+$, so let $x \in \p{x_-, x_+}$. Since $x > x_-$, this means $f\p{x} \geq y$, and because $x < x_+$, $f\p{x} \leq y$, so $f\p{x} = y$, which is a contradiction. Thus, $x = x_- = x_+$, which means $f_-\p{x} \leq y \leq f_+\p{x}$.

Since $y \notin f\p{\R}$, this means that $f_-\p{x} < y$. On the other hand, if $y = f_+\p{x}$, then $f_+\p{x} \notin f\p{\R} \implies I_x = \poc{f_-\p{x}, f_+\p{x}}$. If $y \neq f_+\p{x}$, then $y \in \p{f_-\p{x}, f_+\p{x}} \subseteq I_x$. Hence, $f\p{\R}^\comp = \bigcup_{x \in D} I_x$. Thus, $f\p{\R}^\comp$ is a countable union of Borel sets (intervals) hence Borel, which means that $f\p{\R}$ is also Borel.

Another important property is that $f$ is injective after removing a countable set. Indeed, by monotonicity, $f^{-1}\p{\set{y}}$ is either empty, a singleton, or a non-trivial interval. Let $C$ be the set of $y$ where $f^{-1}\p{\set{y}}$ is an interval. By density, this contains a rational number $r\p{y}$, and if $y \neq z$, then $f^{-1}\p{\set{y}} \cap f^{-1}\p{\set{z}} = \emptyset$, so $r\p{y} \neq r\p{z}$, i.e., $y \mapsto r\p{y}$ is an injection, so $C$ is countable.

Let $\mathcal{F} = \set{A \mid f\p{A} \text{ is Borel}}$. We will show that $\mathcal{F}$ is a $\sigma$-algebra containing the closed intervals.

Given $\br{a, b}$, monotonicity of $f$ implies

which is an intersection of Borel sets, so $\mathcal{F}$ contains closed intervals.

First, it's clear that $\emptyset \in \mathcal{F}$. We also know that $\bigcup_{n=1}^\infty f\p{A_n} = f\p{\bigcup_{n=1}^\infty A_n}$ for any collection of sets. Thus, to show that $\mathcal{F}$ is a $\sigma$-algebra, we just need to show that it is closed under complements. Let $A \in \mathcal{F}$ so that

Notice that if $y \in f\p{A} \cap f\p{A^\comp}$, then there exist $x \neq x'$ so that $f\p{x} = f\p{x'}$, so $f = y$ on an interval containing $x$ and $x'$. Thus, $y \in C$, so $f\p{A} \cap f\p{A^\comp}$ is at most countable which means

is Borel, so $A^\comp \in \mathcal{F}$. Thus, $\mathcal{F}$ is a $\sigma$-algebra containing closed intervals, so $\mathcal{F}$ contains all Borel sets.