Integration with respect to $\nu=\mu_1-\mu_2$

Question

Let $\nu=\mu_1-\mu_2$ be a signed measure written as the difference of two finite positive measures $\mu_1,\mu_2$ on the measure space $(X,\Sigma)$. Let $\nu=\nu^+-\nu^-$ be its Jordan decomposition, let $f:X\to \mathbb R$ be measurable and suppose that $f$ is integrable with respect to $\mu_1,\mu_2$.

Since $\nu^+\leq \mu_1$ and $\nu^-\leq \mu_2$ we have that $f$ is integrable with respect to $\nu^+,\nu^-$. Moreover from

$$\mu_1+\nu^-=\mu_2+\nu^+$$

we get that

$$\int f \,d\mu_1+\int f \,d\nu^-=\int f \,d\mu_2+ \int f \,d\nu^+$$

or $$\int f \,d\nu=\int f \,d\mu_1-\int f \,d\mu_2 \quad \quad (1)$$

Question: Does $(1)$ still holds if we assume $f$ integrable with respect to $\nu^+,\nu^-$?

Thanks a lot for your help.

Answer 1

No. You may end up with $\infty-\infty$. One obvious example is that $\nu=0$. In this case $\nu^+=\nu^-=0$. However, we may choose any finite measure $\mu$ and write $\nu=\mu-\mu$. Choose a non-negative measurable function $f$ such that $\int f du=\infty$...