TITLE: For complex measure $\mu$ and $0< f \le g$ is it true that $\Big|\int_{\Bbb R^n } f d\mu\Big| \le \Big|\int_{\Bbb R^n } g d\mu\Big|$?
QUESTION [5 upvotes]: Let $\mu$ be a complex measure on $\Bbb R^{n}$ and $f,g \in L^1(\mu)$ such that $0< f(x) \le g(x)$ for a.e. $x \in \Bbb R^n$. Then is it true that $$\Big|\int_{\Bbb R^n } f d\mu\Big| \le \Big|\int_{\Bbb R^n } g d\mu\Big| \text{ ? }$$

I was trying to break each integral first in its real and imaginary part and then their corresponding positive and negative parts, i.e. $$\int_{\Bbb R^n } f d\mu = Re\Big(\int_{\Bbb R^n } f d\mu\Big) +i Im \Big(\int_{\Bbb R^n } f d\mu\Big)$$ $$=Re^+\Big(\int_{\Bbb R^n } f d\mu\Big)-Re^-\Big(\int_{\Bbb R^n } f d\mu\Big)+iIm^+\Big(\int_{\Bbb R^n } f d\mu\Big)-iIm^-\Big(\int_{\Bbb R^n } f d\mu\Big)$$ and then wanted to look at corresponding decompositions of the measure and the integral $\int_{\Bbb R^n } g d\mu$ and obtain inequalities. But I can't figure them out and proceed from here.
Can someone please help?

REPLY [3 votes]: Hint: If $\mu =\delta_0-\delta_1$ then the inequality becomes $|f(0)-f(1)| \leq |g(0)-g(1)|$. Can you come with an example where $0< f \leq g$ but this inequality fails? [ Make RHS $0$].
[Notation: $\delta_x (E)=1$ if $x \in E$ and $0$ otherwise].