= Solution
{c}
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 get
$$
\mathbb E[X\vee(-a)\mid\mathcal G]=Y\vee(-a).
$$
For every real $t$,
$$
(t\wedge a)\vee(-a)=(t\wedge a)+(t\vee(-a))-t.
$$
Using this identity, part (ii), the lower truncation identity, and linearity of <conditional expectation> gives
$$
\boxed{\mathbb E[(X\wedge a)\vee(-a)\mid\mathcal G]
=(Y\wedge a)\vee(-a)\quad\text{almost surely}.}
$$
In 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 $a=1,2,\ldots$ and discard the <countable union> of exceptional <null sets>. At every remaining outcome, let $a\to\infty$ to obtain
$$
\boxed{X=Y\quad\text{almost surely}.}
$$
Thus <conditional expectation preserving a distribution> forces equality even when only <integrability> is assumed.
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