Let $\pi$ be the law of a standard random variable and let $f$ be a measurable function. For a fixed $p$, how is $\pi(f>p)$ computed?
Is this as simple as $A:=\{x : f(x)>p\}$ and $$\pi(f>p) = \int_{A}\pi(dx)= \frac{1}{\sqrt{2\pi}} \int_{A} e^{-x^2/2} dx.$$
I'm not sure if $f$ plays any other role in the integral.