Prove that if $|f|$ is gauge integrable then so is $f$.

Question

I have tried to prove this using the Cauchy criterion for HK-integration but I have been unsuccessful thus far. I have various fancy theorems at my disposal such the MCT and the DCT but I can not think of a way to prove this seemingly basic result. If someone knows a way to prove this please let me know. I don't need an entire proof just something to get me started and I can fill out the details.

Perhaps my claim is wrong, in which case a counter-example is sufficient.

Answer 1

If we allow the existence of non-measurable sets (and hence the axiom of choice) it's not true. Pick any non-measurable set $V \subset [0,1]$ and define $$f = \begin{cases} -1 & x \in V\\ 1 & x\in[0,1]\cap V^c \\ 0 & \textrm{otherwise}\end{cases}$$ Then $f$ is bounded and has compact support, so HK integrability and Lebesgue integrability are equivalent. Evidently $\lvert f\rvert$ is integrable, while $f$ is not.