I'm studying measure theory and two question came to my mind:
- If $f:\mathbb{R}\to\mathbb{R}$ is monotone and $B\subseteq\mathbb{R}$ is borel, is the image $f(B)$ borel?
- 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.
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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).
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If this were true, maybe we could use some convergence argument to solve the problem above.
Best Answer
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.