= Conditional Markov inequality with a measurable threshold
{title2=$\mathbb P(|X|\geq Z\mid\mathcal G)\leq Z^{-1}\mathbb E[|X|\mid\mathcal G]$}
If $X$ is an <integrable random variable> and $Z$ is a finite, strictly positive $\mathcal G$-<measurable> <random variable>, then
$$
\mathbb E[\mathbf1_{\{|X|\geq Z\}}\mid\mathcal G]
\leq\frac{\mathbb E[|X|\mid\mathcal G]}{Z}\quad\text{a.s.}
$$
Indeed, $0\leq Z\mathbf1_{\{|X|\geq Z\}}\leq|X|$. Monotonicity of <conditional expectation> and the <measurable-factor rule for conditional expectation> give
$$
Z\mathbb E[\mathbf1_{\{|X|\geq Z\}}\mid\mathcal G]
\leq\mathbb E[|X|\mid\mathcal G].
$$
Division 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.
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