Monotone functions and Borel sets

Question

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.

Answer 1

The answer to #1 is yes.

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.

Answer 2

Here's a self-contained answer to #1.

First note/recall that any union of half-open intervals of the form $[a, b)$ is Borel. To see this, suppose that we have such a collection $[a_{x}, b_{x})$ for $x$ in some index set $X$. The set $U = \bigcup_{x\in X}(a_{x}, b_{x})$ is open, so it suffices to see that the set $Y = \{ a_{x} : x \in X\} \setminus U$ is countable. To see that $Y$ is countable, note that we can pick for each $a \in Y$ a rational number $q_{a}$ which is in $(a_{x}, b_{x})$ for some $x \in X$ with $a_{x} = a$. Then if $x, y$ in $X$ are such that $a_{x}, a_{y}$ are in $Y$ with $a_{x} < a_{y}$, $q_{a_{x}} < b_{x}$ and $q_{a_{y}} < b_{y}$, it must be that $b_{x} \leq a_{y}$, which means that $q_{a_x} < b_{x} \leq a_{y} < q_{a_y}$.

It follows that that any union of half-open intervals is Borel.

Now let $B \subseteq \mathbb{R}$ be Borel and let $f \colon B \to \mathbb{R}$ be an increasing function. Let $E$ be the set of $x \in \mathbb{R}$ for which the set $\{ y \in B : f(y) \leq x\}$ is nonempty and bounded above. If $B$ is bounded above then $E$ is the union of all intervals of the form $[f(y), \infty)$ for $y \in B$. Otherwise, $E$ is the union of all intervals of the form $[f(y), f(z))$ for all $y \leq z$, both in $B$. Either way, $E$ is Borel.

Define $g$ on $E$ by letting $g(x) = \sup\{ y\in B : f(y) \leq x\}$. Then $g$ is an increasing function with Borel domain, so Borel, so $g^{-1}[B]$ is Borel. To see that $f[B]$ is Borel, it suffices to see that $g^{-1}[B] \setminus f[B]$ and $f[B] \setminus g^{-1}[B]$ are both Borel.

To see the former, suppose that $x$ is in $g^{-1}[B] \setminus f[B]$. Then there exists a $z \in B$ such that $\sup\{ y\in B : f(y) \leq x\} = z$, but $f(z) \neq x$. If $f(z) < x$, then everything in the interval $(f(z), x]$ is in $g^{-1}[B] \setminus f[B]$. If $f(z) > x$, then everything in the interval $[x, f(z))$ is in $g^{-1}[B] \setminus f[B]$. This shows that $g^{-1}[B] \setminus f[B]$ is a union of half-open intervals.

To see the latter, we show that $f[B] \setminus g^{-1}[B]$ is countable. If $x \in f[B] \setminus g^{-1}[B]$, then there is a $z \in B$ such that $f(z) = x$, but $g(x) \neq z$, which means that there is a $w > z$ in $B$ such that $f(w) = x$. So the members of $f[B] \setminus g^{-1}[B]$ induce a collection of pairwise disjoint nonempty open intervals $(z, w)$, which means that there are only countably many of them.