Theorem 1   Suppose that $\mathbb{E}[X^2]<\infty$ . Then there is a $\mathcal{G}$ -measurable random variable $Y$ satisfying $\mathbb{E}[Y^2]<\infty$ and equation () is satisfied for all $G\in\mathcal{G}$ .

Proof. Consider the norm $\Vert Y\Vert_2\equiv\mathbb{E}[Y^2]^{1/2}$ on the vector space $V=L^2(\Omega,\mathcal{F},\mathbb{P})$ of real valued random variables $Y$ satisfying $\mathbb{E}[Y^2]<\infty$ (up to $\mathbb{P}$ almost everywhere equivalence). This is given by the following inner product \begin{equation*} \langle Y_1,Y_2\rangle\equiv\mathbb{E}[Y_1Y_2]. \end{equation*}As $L^p$ -spaces are complete, this makes $V$ into a Hilbert space (see also, $L^2$ -spaces are Hilbert spaces). Then, $U\equiv L^2(\Omega,\mathcal{G},\mathbb{P})$ is a complete, and hence closed, subspace of $V$ .

By the existence of orthogonal projections onto closed subspaces of Hilbert spaces, there is an orthogonal projection $\pi\colon V\rightarrow U$ . In particular, $\langle\pi Y-Y,Z\rangle=0$ for all $Y\in V$ and $Z\in U$ . Setting $Y=\pi X$ gives \begin{equation*} \mathbb{E}[1_GY]-\mathbb{E}[1_GX] = \langle 1_G,\pi X-X\rangle=0 \end{equation*}as required.

Theorem 2   Let $X$ be a nonnegative random variable taking values in $\mathbb{R}\cup\{\infty\}$ . Then, there exists a nonnegative $\mathcal{G}$ -measurable random variable $Y$ taking values in $\mathbb{R}\cup\{\infty\}$ and satisfying () for all $G\in\mathcal{G}$ . Furthermore, $Y$ is uniquely defined $\mathbb{P}$ -almost everywhere.

Proof. First, let $X_n=\min(n,X)$ . As this is bounded, theorem 1 says that the conditional expectations $Y_n=\mathbb{E}[Y_n\mid\mathcal{G}]$ exist. Furthermore, as $X_0=0$ , we may take $Y_0=0$ . For any $n$ , setting $G=\{Y_{n+1}<Y_n\}\in\mathcal{G}$ gives \begin{equation*} \mathbb{E}[1_G(Y_n-Y_{n+1})]=\mathbb{E}[1_G(X_n-X_{n+1})]\le 0. \end{equation*}So $1_G(Y_n-Y_{n+1})$ is a nonnegative random variable with nonpositive expectation, hence is almost surely equal to zero. Therefore, $Y_{n+1}\ge Y_n$ (almost surely) and, by replacing $Y_n$ with the maximum of $Y_1,\ldots\,Y_n$ we may suppose that $(Y_n)$ is an increasing sequence of random variables. Setting $Y=\sup_nY_n$ , the monotone convergence theorem gives \begin{equation*} \mathbb{E}[1_GY]=\lim_{n\rightarrow\infty}\mathbb{E}[1_GY_n]=\lim_{n\rightarrow\infty}\mathbb{E}[1_GX_n]=\mathbb{E}[1_GX] \end{equation*}as required.

Finally, suppose that $\tilde Y$ is also a nonnegative $\mathcal{G}$ -measurable random variable satisfying (). For any $x\in\mathbb{R}$ , setting $G=\{\tilde Y>Y,x>Y\}$ then $1_GY$ is bounded and, \begin{equation*} \mathbb{E}[1_G(\tilde Y-Y)]=\mathbb{E}[1_GX]-\mathbb{E}[1_GX]=0 \end{equation*}showing that $\mathbb{P}(G)=0$ . Letting $x$ increase to infinity gives $\tilde Y\le Y$ (almost surely) and, similarly, $Y\le \tilde Y$ so that $Y=\tilde Y$ almost surely.

Theorem 3   Let $X$ be a random variable such that $\mathbb{E}[|X|\mid\mathcal{G}]<\infty$ almost surely. Then, there exists a $\mathcal{G}$ -measurable random variable $Y$ such that $\mathbb{E}[1_G|Y|]<\infty$ and () is satisfied for every $G\in\mathcal{G}$ with $\mathbb{E}[1_G|X|]<\infty$ .

Furthermore, $Y$ is uniquely defined up to $\mathbb{P}$ -a.e. equivalence.

Proof. The positive and negative parts $X_+,X_-$ of $X$ satisfy \begin{equation*} \mathbb{E}[X_+\mid\mathcal{G}]+\mathbb{E}[X_-\mid\mathcal{G}]=\mathbb{E}[|X|\mid\mathcal{G}]<\infty \end{equation*}almost surely. We can therefore set $Y_{\pm}\equiv\mathbb{E}[X_\pm\mid\mathcal{G}]$ and $Y=Y_+-Y_-$ .

If $G\in\mathcal{G}$ satisfies $\mathbb{E}[1_G|X|]<\infty$ then $\mathbb{E}[1_GY_\pm]=\mathbb{E}[1_GX_\pm]<\infty$ , so $\mathbb{E}[1_G|Y|]<\infty$ and, \begin{equation*} \mathbb{E}[1_GY]=\mathbb{E}[1_GY_+]-\mathbb{E}[1_GY_-]=\mathbb{E}[1_GX_+]-\mathbb{E}[1_GX_-]=\mathbb{E}[1_GX] \end{equation*}as required.

Finally, suppose that $\tilde Y$ satisfies the same conditions as $Y$ . For any $x\ge 0$ set $G=\{Y_++Y_-\le x,\tilde Y> Y\}\in\mathcal{G}$ . Then, \begin{equation*} \mathbb{E}[1_G|X|]=\mathbb{E}[1_G(Y_++Y_-)]\le x<\infty. \end{equation*}So, $\mathbb{E}[1_G|Y|]$ and $\mathbb{E}[1_G|\tilde Y|]$ are finite, hence () gives \begin{equation*} \mathbb{E}[1_G(\tilde Y - Y)]=\mathbb{E}[1_GX]-\mathbb{E}[1_GX]=0. \end{equation*}So $\mathbb{P}(G)=0$ and, letting $x$ increase to infinity, $\tilde Y\le Y$ almost surely. Similarly, $Y\le\tilde Y$ and therefore $\tilde Y=Y$ almost surely.