= Solution
Put $A=\{|X|\geq Z\}$. Both $A$ and its <indicator function> are <measurable>, and $0\leq Z\mathbf1_A\leq |X|$. Thus $Z\mathbf1_A$ is an <integrable random variable>. Monotonicity of <conditional expectation> gives
$$
\mathbb E[Z\mathbf1_A\mid\mathcal G]\leq\mathbb E[|X|\mid\mathcal G]\quad\text{a.s.}
$$
The measurable-factor property of <conditional expectation> says that a bounded $\mathcal G$-<measurable function> $H$ satisfies $\mathbb E[HW\mid\mathcal G]=H\mathbb E[W\mid\mathcal G]$ for every <integrable random variable> $W$. Here the factor $Z$ need not be bounded. Apply that property first to $Z\wedge m$ and $W=\mathbf1_A$, then use <monotone convergence> on the nonnegative products, including the conditional version, to obtain
$$
\mathbb E[Z\mathbf1_A\mid\mathcal G]
=\lim_{m\to\infty}(Z\wedge m)\mathbb E[\mathbf1_A\mid\mathcal G]
=Z\mathbb E[\mathbf1_A\mid\mathcal G].
$$
All the left-hand <conditional expectations> are well defined; the final one is finite <almost surely> because $Z\mathbf1_A$ is an <integrable random variable>. Since $Z>0$ <almost surely>, division now yields
$$
\boxed{\mathbb E[\mathbf1_{\{|X|\geq Z\}}\mid\mathcal G]
\leq Z^{-1}\mathbb E[|X|\mid\mathcal G]\quad\text{a.s.}}
$$
This proves the <conditional Markov inequality with a measurable threshold> without assuming that $Z^{-1}|X|$ is an <integrable random variable>. The measurable-factor rule, with its product-integrability condition, is also stated in the standard <conditional expectation> theory.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[§4.1.2, Theorem 4.1.14, printed pp. 212–213: bringing out a measurable factor]{ref}
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