landdenaw

2022-06-28

Let $f:X\to \mathbb{R}$ be a continuous function on a compact metric space $X$. Assume that a Borel probability measure $\mu$ is absolutely continuous with respect to Lebesgue measure Leb.
Is it true that if $f\left(x\right)<0$ for Leb a.e. x, then $\int fd\mu <0?$?
I think it should be true as $\mu <<\text{Leb}$.
Attempt: Assume that $\int fd\mu \ge 0$. Then $f\ge 0$ μ a.e. x. That means $\mu \left(\left\{x:f\left(x\right)>0\right\}\right)>0$. That implies $\text{Leb}\left(\left(\left\{x:f\left(x\right)>0\right\}\right)>0$ which is not true.

klemmepk

Expert

The compactness, metrizability and other conditions are unnecessary. All that is needed is that f is measurable and $\mu \ll \lambda$, where I use $\lambda$ for Lebesgue measure.
$\lambda \left\{x\in X:f\left(x\right)\ge 0\right\}=0\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}\mu \ll \lambda \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mu \left\{x\in X:f\left(x\right)\ge 0\right\}=0$
And immediately we have that, as X is not $\mu$-null:
${\int }_{X}f\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mu <0$
I'm afraid we do not quite have:

Since a function that is both positive and negative can simply cancel, e.g.
${\int }_{\mathbb{R}}\frac{\mathrm{sin}x}{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$
Famously evaluates to $\pi >0$, but of course $\frac{\mathrm{sin}\left(x\right)}{x}$ is negative on a set of infinite measure.
Otherwise, what you said is fine.

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