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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 29 / 1 / b / Solution

 Solution (source code)