The standard proof that the sum of measurable functions is measurable uses countable choice, via the countable subadditivity of outer measure ($\implies$ measurable sets are closed under countable union). I've been trying to puzzle out whether this usage is required, using a few standard counterexample models:
- It is consistent with $\sf ZF$ that $\Bbb R=\bigcup_{n\in\Bbb N}R_n$ is a countable union of countable sets.
- It is consistent with $\sf ZF$ that there is a Lebesgue nonmeasurable set $A=\bigcup_{n\in\Bbb N}A_n$ that is a countable union of countable sets.
(I actually made the second statement up, but given the results in this area it seems plausible. It is of course sufficient to have a model of the first statement, in which there exists a nonmeasurable set, to satisfy the second.)
Clearly, countable subadditivity of $\lambda^*$ is violated in these models: compare $\lambda^*(\bigcup_{n\in\Bbb N}R_n)=\infty$ to $\sum_{n\in\Bbb N}\lambda^*(R_n)=0$, where the $R_n$ are assumed disjoint. Measurable sets are also not closed under countable union in the second model, with $A$ as the counterexample.
What I haven't managed to show is whether it is false that measurable functions are closed under addition. (Here a measurable function is one such that $f^{-1}(O)$ is measurable for every open set $O$, not just the intervals.) Is there a function $f(x)$ defined in terms of the $A_n$'s such that $f(x)+x$ or something similar is not measurable?
I would also be interested in the following special case: A function of bounded variation / difference of two monotone functions is measurable.
Best Answer
For your last Q. If $f:r\to R$ is increasing and $r\in R,$ then $f^{-1}(r,\infty)$ is $\phi, $ or is $R, $ or is $(\;\inf f^{-1}(r,\infty)\;), $ or is $[\min f^{-1}\{r\},\infty\;), $ which is in all cases a Borel set. Similarly, for $s\in R, $ the set $f^{-1}(-\infty,s)$ is Borel. So $f^{-1}(s,r)$ is Borel. Every open set of reals is a countable union of open intervals (see below); therefore $f^{-1}U$ is Borel for every open $U.$ All of this can be done in ZF.
To show in ZF that every open $U\subset R$ is a countable union of pair-wise disjoint open intervals (including possibly unbounded ones): For convenience let In $(x,y)=(x,y)\cup (y,x)$ for $x,y \in R.$ (In $(x,y)$ is the open interval between $x$ and $y.$) For any open $U\subset R$ and $x,y\in U$ let $x\equiv_U y\iff$ In$(x,y)\subset U.$ We easily show this is an equivalence relation on $U,$ and that each equivalence class $[x]_{\equiv_U}$ is an open interval. Now $U=\cup \{[x]_{\equiv_U} :x\in Q\cap V\}.$