 Conditional Markov inequality with a measurable threshold

If $X$ is an integrable random variable and $Z$ is a finite, strictly positive $\mathcal{G}$-measurable random variable, thenIndeed, $0\le Z{1}_{\{|X|\ge Z\}}\le |X|$. Monotonicity of conditional expectation and the measurable-factor rule for conditional expectation giveDivision by the strictly positive $Z$ proves the claim. Only the displayed product needs to be an integrable random variable; neither $Z$ nor $|X|/Z$ needs to have finite expected value. The measurable threshold represents information already available before asking how likely the remaining exceedance is.