Conditional Expectation

Take first $X\ge 0$. Define a measure on $(\Omega,\G)$ by $\nu(G)=\int_G X\dd\P$. It is finite, with $\nu(\Omega)=\E[X]<\infty$, and absolutely continuous with respect to the restriction $\P\!\restriction_\G$, since $G\in\G$ with $\P(G)=0$ forces $\int_G X\dd\P=0$. The Radon-Nikodym theorem applied on $(\Omega,\G,\P\!\restriction_\G)$ supplies a $\G$-measurable $Y\ge 0$ with $\nu(G)=\int_G Y\dd\P$ for all $G\in\G$, which is exactly Equation (1). For general $X$ set $\E[X\mid\G]=\E[X^+\mid\G]-\E[X^-\mid\G]$, both terms finite since $X^\pm\le\abs X\in L^1$. For uniqueness, if $Y_1,Y_2$ both satisfy Equation (1), the $\G$-set $G=\{Y_1>Y_2\}$ gives $\int_G(Y_1-Y_2)\dd\P=0$ with a nonnegative integrand, so $\P(G)=0$, and symmetrically, hence $Y_1=Y_2$ almost surely.

The space $L^2(\Omega,\G,\P)$ is complete, being $L^2$ of the measure space $(\Omega,\G,\P\!\restriction_\G)$, hence a closed subspace of $L^2(\P)$ (when $\G$ is not $\P$-complete one works with the completion $\bar\G$, and $L^2(\bar\G)=L^2(\G)$ inside $L^2(\P)$ since each $\bar\G$-measurable function agrees a.s. with a $\G$-measurable one). So the projection $Y$ exists and is characterized by $X-Y\perp L^2(\G)$, that is $\E[(X-Y)Z]=0$ for every $Z\in L^2(\G)$. In particular for $Z=\ind_G$, bounded and hence in $L^2$ on a finite measure space, this recovers Equation (1), so $Y=\E[X\mid\G]$ by Theorem 2. The minimization statement follows because an orthogonal projection minimizes the distance from $X$ to the subspace.

Statement (i) holds because the right side is $\G$-measurable and integrates correctly over each $G\in\G$ by linearity of the integral, so Theorem 2 identifies it. For (ii), the inner variable $W=\E[X\mid\G]$ lies in $L^1$, so $\E[W\mid\mathcal H]$ is defined. The right side $\E[X\mid\mathcal H]$ is $\mathcal H$-measurable, and for $H\in\mathcal H\subseteq\G$, since $H\in\G$ as well, $\int_H\E[X\mid\mathcal H]\dd\P=\int_H X\dd\P=\int_H W\dd\P$ by Equation (1) applied at each level. By uniqueness Theorem 2 identifies $\E[X\mid\mathcal H]$ with $\E[W\mid\mathcal H]$. For (iii), the claim holds for $Z=\ind_{G_0}$ with $G_0\in\G$ because for $G\in\G$,

and it extends to general bounded $\G$-measurable $Z$ by dominated convergence. In detail, with $\abs Z\le M$ pick simple $\G$-measurable $Z_n$ with $\abs{Z_n}\le M$ and $Z_n\to Z$ pointwise; then $\abs{Z_nX}\le M\abs X\in L^1$ and $Z_nX\to ZX$ almost surely, so the ordinary dominated convergence theorem gives $\|Z_nX-ZX\|_1\to0$, and since conditional expectation is an $L^1$-contraction we get $\|\E[Z_nX\mid\G]-\E[ZX\mid\G]\|_1\le\|Z_nX-ZX\|_1\to0$. Hence $\E[Z_nX\mid\G]\to\E[ZX\mid\G]$ in $L^1$, so along a subsequence almost surely, while $Z_n\E[X\mid\G]\to Z\E[X\mid\G]$ almost surely; equating limits gives the claim. For (iv), if $X\ge0$ a.s. set $G=\{\E[X\mid\G]<0\}\in\G$; then $\int_G\E[X\mid\G]\dd\P=\int_G X\dd\P\ge0$ while the integrand is negative on $G$, forcing $\P(G)=0$.

At each rational $q$ a convex $\varphi$ has a subgradient $s_q$, giving the affine minorant $\ell_q(x)=\varphi(q)+s_q(x-q)\le\varphi(x)$; continuity of $\varphi$ and density of $\Q$ then yield $\sup_q\ell_q(x)=\varphi(x)$ for every $x\in\R$, a supremum over a countable family. For each $q$, monotonicity (iv) and linearity (i) from Proposition 4, together with $\E[b\mid\G]=b$ for the constant $b$ (immediate from Theorem 2, since a constant is $\G$-measurable and reproduces its own averages), give $\E[\varphi(X)\mid\G]\ge\E[\ell_q(X)\mid\G]=s_q\E[X\mid\G]+(\varphi(q)-s_q q)=\ell_q\!\big(\E[X\mid\G]\big)$ almost surely. Taking the supremum over the countable family, a null set at a time, yields $\E[\varphi(X)\mid\G]\ge\sup_q\ell_q\!\big(\E[X\mid\G]\big)=\varphi\!\big(\E[X\mid\G]\big)$ [1].