= Measurable-factor rule for conditional expectation
{title2=$\mathbb E[HY\mid\mathcal G]=H\mathbb E[Y\mid\mathcal G]$}
Let $Y$ be an <integrable random variable>, let $H$ be a finite $\mathcal G$-<measurable> <random variable>, and suppose $HY$ is an <integrable random variable>. Then a known factor can be taken outside the <conditional expectation>:
$$
\mathbb E[HY\mid\mathcal G]=H\mathbb E[Y\mid\mathcal G]\quad\text{a.s.}
$$
For a bounded $H$, first verify the defining <integral> identity when $H$ is the <indicator function> of a <measurable set> in $\mathcal G$, extend by <linearity> to <simple functions>, and then use bounded approximation and the <dominated convergence theorem>. For nonnegative $H,Y$, truncating $H$ and using the <conditional monotone convergence theorem> proves the identity with extended nonnegative <conditional expectations>. In particular,
$$
\mathbb E\bigl[|H|\mathbb E[|Y|\mid\mathcal G]\bigr]=\mathbb E|HY|<\infty.
$$
For general $H,Y$ under the stated <integrability> assumptions, the <conditional Jensen inequality> bounds $|H\mathbb E[Y\mid\mathcal G]|$ by that <integrable random variable>. Apply the bounded-factor result to truncations of $H$, then use the <dominated convergence theorem> on the products and the <L1 contraction of conditional expectation>. This proves the identity without assuming $H$ itself is an <integrable random variable>.
The product-integrability condition is essential to the ordinary finite <conditional expectation> statement. An unbounded factor cannot be moved through an <expected value> simply because its partner is an <integrable random variable>.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[§4.1.2, Theorem 4.1.14, printed pp. 212–213]{ref}
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