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Again put $Y=\mathbb{E}[X\mid \mathcal{G}]$. For any fixed real $c$, the random variableis nonnegative and integrable. The sign function of $Y-c$ is bounded and $\mathcal{G}$-measurable, so conditioning and equality of the probability distributions giveThus $D_c=0$ almost surely. On $\{Y>c\}$ this forces $X\ge c$; on $\{Y<c\}$ it forces $X\le c$; on $\{Y=c\}$ it forces $X=c$. In particular $\{Y=c\}\subseteq \{X=c\}$ up to a null set. These two events have equal probability, again by equality of the probability distributions, so they agree up to a null set. ConsequentlyTaking $c=0$ establishes the suggested sign assertion, but this alone would not exclude unequal positive values. Apply the same argument simultaneously to the countable set of rational $c$. Outside the union of their null sets, unequal $X$ and $Y$ would have a rational strictly between them, giving opposite signs. This proves the conditional expectation preserving a distribution result:Only integrability was used; no finite second moment or hidden square-integrability assumption is needed.