Taking out what is known

Write $U=E[X\mid \mathcal{G}]$. If $Z={\mathbf{1}}_B$ with $B\in \mathcal{G}$, then for every $A\in \mathcal{G}$, ${\int }_{A}ZU={\int }_{A\cap B}U={\int }_{A\cap B}X={\int }_{A}ZX$. The products are integrable and $ZU$ is $\mathcal{G}$-measurable, so uniqueness gives the identity. Finite sums of such indicators give the identity for bounded simple $Z$ by linearity.

If $\mathrm{\mid }Z\mathrm{\mid }\le M$, apply [F4] to $Z^{+}$ and $Z^{-}$ and subtract their approximations to obtain simple $Z_n\to Z$ with $\mathrm{\mid }{Z}_n\mathrm{\mid }\le M$. Then $\mathrm{\mid }{Z}_{n}X\mathrm{\mid }\le M\mathrm{\mid }X\mathrm{\mid }$ and $\mathrm{\mid }{Z}_{n}U\mathrm{\mid }\le M\mathrm{\mid }U\mathrm{\mid }$, both integrable majorants. By [F5] in each event identity of step 1.1, ${\int }_{A}ZU={\int }_{A}ZX$. The limit $ZU$ is measurable and integrable, so [F3] proves the bounded case.

For finite measurable $Z$ with $ZX\in L^1$, set $V=E[\mathrm{\mid }X\mathrm{\mid }\mid \mathcal{G}]\ge 0$ and $R_n=\mathrm{\mid }Z\mathrm{\mid }\wedge n$. The bounded case and expectation preservation imply $E[R_{n}V]=E[R_n\mathrm{\mid }X\mathrm{\mid }]$. MCT gives $E[\mathrm{\mid }Z\mathrm{\mid }V]=E\mathrm{\mid }ZX\mathrm{\mid }<\mathrm{\infty }$. Since $\mathrm{\mid }U\mathrm{\mid }\le V$ almost surely, $E\mathrm{\mid }ZU\mathrm{\mid }\le E\mathrm{\mid }ZX\mathrm{\mid }<\mathrm{\infty }$.

Let $Z_n=\max (-n,\min (Z,n))$. For each event $A$ the bounded case gives ${\int }_A{Z}_{n}U={\int }_A{Z}_{n}X$. Now $Z_{n}U\to ZU$ and $Z_{n}X\to ZX$, dominated by the integrable $\mathrm{\mid }ZU\mathrm{\mid }$ and $\mathrm{\mid }ZX\mathrm{\mid }$ respectively. DCT and uniqueness yield $E[ZX\mid \mathcal{G}]=ZU$. This proves both the stated two-product extension and its stronger integrability observation.

The same indicator identity gives locality: if $X=Y$ almost surely on $B\in \mathcal{G}$, then ${\mathbf{1}}_{B}E[X\mid \mathcal{G}]=E[{\mathbf{1}}_{B}X\mid \mathcal{G}]=E[{\mathbf{1}}_{B}Y\mid \mathcal{G}]={\mathbf{1}}_{B}E[Y\mid \mathcal{G}]$ as classes. No assertion is made about arbitrary values on exceptional points.