I'm investigating the properties of Lebesgue measure without countable additivity. That is, I have an outer measure on $\Bbb R$ that satisfies (for subsets $A,B$):
- $m^*(A)\in[0,\infty]$
- $m^*(A\cup B)\le m(A)+m(B)$
- $m^*(A)=0$ when $A$ is countable
- $m^*(A)\le m^*(B)$ when $A\subseteq B$
- $m^*(A+x)=m^*(A)$
- $m^*([a,b])=b-a$
The actual measure $m(A)$ is as usual defined to be equal to $m^*(A)$ when $m(A)=m(A\cap E)+m(A\setminus E)$ for all sets $E$ (in which case $A$ is called measurable). From this it is derived that:
- $A$ is measurable when $m^*(A)=0$
- $A\cup B$ and $A^c$ are measurable if $A,B$ are
- intervals are measurable
Notably absent from this list are theorems concerning countable unions and intersections, except, curiously, that countable sets are still null. Other than that, it seems to act a lot like Jordan measure. Given this setting, I want to define measurable functions analogous to the standard definition: a function $f$ is measurable when $f^{-1}((a,\infty))$ is measurable for each $a\in\bar{\Bbb R}$.
But the snag here is that most proofs need countable additivity for even the most basic properties, and the textbook definition of measure usually includes countable additivity, so I have no references. In particular, I would like to know if $f+g$ and $-f$ are measurable given that $f$ and $g$ are. The usual proof uses $(-\infty,a)=\bigcup_{n\in\Bbb N}\left(\Bbb R\setminus(a-\frac1n,\infty)\right)$, so that won't work for me. I am a little blind because I have no counterexamples in mind, so any help with a go-to counterexample here would be great too.