 Solution

Equality of the probability distributions implies equality of all integrals for which they are finite, in particular $\mathbb{E}[(\mathbb{E}[X\mid \mathcal{G}]{)}^2]=\mathbb{E}[X^2]<\infty$. Part (a), applied to this conditional expectation, therefore gives $\boxed{X=\mathbb{E}[X\mid \mathcal{G}]\text{ almost surely}.}$ The second-moment assumption justifies the squared-norm argument; the next part removes it.