# Monotone functions and Borel sets

I'm studying measure theory and two question came to my mind:

1. If $f:\mathbb{R}\to\mathbb{R}$ is monotone and $B\subseteq\mathbb{R}$ is borel, is the image $f(B)$ borel?
2. If $f:\mathbb{R}\to\mathbb{R}$ is a monotone function (say, non-decreasing), does there exist a sequence of continuous functions $f_n:\mathbb{R}\to\mathbb{R}$ converging pointwise to $f$?

Here's the motivation for those questions: Let $(X,M)$ and $(Y,N)$ be measure spaces.

1. It's known that if $\mu$ is a measure on $(X,M)$ and $f:X\to Y$ is measure, then we have the pushforward measure $f_*\mu(A)=\mu(f^{-1}(A))$ on $(Y,N)$. What if we were to define a "pullback measure"? Given a function $f:X\to Y$ such that $f(M)\subseteq N$ (i.e. $f$ maps measurable sets to measurable sets) and a measure $\nu$ on $N$, the natural formula would be $f^*\nu(A)=\nu(f(A))$. So we ask if is there a good amount of functions which map measurable sets to measurable sets in $\mathbb{R}$, and monotone functions seem like good candidates (for strictly monotone, continuous functions, the result is valid and there are several answers on the web).

2. If this were true, maybe we could use some convergence argument to solve the problem above.

• Your attempted definition does not work: $f^*\nu$ is not additive in general. – user138530 Oct 4 '14 at 5:10
• @ChristianRemling I see that. In that paragraph, I'm simply explaining how I got to question 1. – Questioner Oct 4 '14 at 5:19

First note that if $f$ is monotone, it is Borel. (The sets $(a, \infty)$ generate the Borel $\sigma$-algebra, and $f^{-1}((a, \infty))$ is Borel for each $a$ because it is of the form $(b, \infty)$ or $[b, \infty)$ (for increasing functions) or $(-\infty, b)$ or $(-\infty, b]$ (for decreasing functions).)
Now for each $y$, $f^{-1}(\{y\})$ is either empty, a point, or a nontrivial interval. Let $C$ be the set of all $y$ such that $f^{-1}(y)$ is a nontrivial interval. Since each interval contains a rational, $C$ is countable. Let $D = f^{-1}(C)$; note that $D$ is Borel.
If $B$ is an arbitrary Borel set, we have $f(B) = f(B \cap D) \cup f(B \setminus D)$. Now $f(B \cap D) \subset C$, hence it is countable and hence Borel. So it suffices to show that $f(B \setminus D)$ is Borel.
On $D^c$, and hence on $B \setminus D$, $f$ is injective. Now it is a theorem of descriptive set theory that an injective Borel function on a Borel subset of a Polish space has a Borel image. (See for instance Theorem 4.5.4 of Srivastava, A Course on Borel Sets.) But $B \setminus D$ is Borel, so $f(B \setminus D)$ is also Borel and we are done.