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Let $T$ be a measure-preserving transformation of a probability space and $X$ an integrable random variable. If the displayed duality holds for every bounded random variable $Z$, test it on indicator functions to identify $\mathbb{E}[X\mid T^{-1}(\mathcal{F})]=X\circ T$. The measure-preserving transformation gives the same probability distribution to $X$ and $X\circ T$. The principle of conditional expectation preserving a distribution then forces equality almost surely. This does not require an inverse for $T$ or require $X$ to be square-integrable.

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.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.