Conditional expectation is the l2 orthogonal projection

The inclusion sends the class of a $\mathcal{G}$-measurable function to its ambient class. Two such functions agree almost surely for the restricted measure exactly when they do for P; their squared integrals are identical. Thus inclusion is well defined, injective, linear and isometric. If a sequence in its image converges in ambient $L^2$, its preimages are Cauchy, converge by [F2] under [F3], and their images converge to the same ambient limit by the isometry and uniqueness of metric limits. Hence the image is closed.

By [F1], $U\in L^2(\mathcal{G})$. Fix $Z\in L^2(\mathcal{G})$. Both $ZX$ and $ZU$ are integrable by [F4]. Taking-out [F5] gives $E[ZX\mid \mathcal{G}]=ZU$. Taking ordinary expectations by [F7] yields $E[ZX]=E[ZU]$, hence $E[Z(X-U)]=0$. This establishes orthogonality for every Z directly, and in particular for bounded G-measurable tests, without a density argument.

For every $Z\in L^2(\mathcal{G})$, expand $X-Z=(X-U)+(U-Z)$. All products are integrable by [F4], and step 1.2 annihilates the cross term. Therefore $E[(X-Z{)}^2]=E[(X-U{)}^2]+E[(U-Z{)}^2]$. The last term is nonnegative and is zero exactly when $U=Z$ as an $L^2$ class, since $L^2$ is a normed space. The minimizer is therefore unique. Finally [F6] fixes every member of the subspace, so the conditional map is indeed the projection onto it.