Proof of Levy's Upward Theorem in Mean Square

Fix conditional expectations $Y_n$ of $X$ given $\mathcal{G}_n$ for each $n\in\mathbb{N}$ and a conditional expectation $Y_\infty$ of $X$ given $\mathcal{G}_\infty$, as in Conditional Expectation of a Square-Integrable Random Variable. All norms, inner products, and expectations below are those of Square-Integrable Random Variables and the Mean-Square Inner Product.

Step 1: The sequence $(Y_n)$ is Cauchy in mean square. Let $m\le n$. Since $\mathcal{G}_m\subseteq\mathcal{G}_n$, both $Y_m$ and $Y_n$ are $\mathcal{G}_n$-measurable and square-integrable, hence so is $Z=Y_n-Y_m$ by the closure properties recorded in Square-Integrable Random Variables and the Mean-Square Inner Product. The orthogonality property (property 2 of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, which every conditional expectation satisfies by the equivalence stated there) gives $\mathbb{E}[(X-Y_n)Z]=0$. Writing $X-Y_m=(X-Y_n)+Z$ and expanding with the bilinearity of the mean-square inner product from Square-Integrable Random Variables and the Mean-Square Inner Product,

Set $d_n=\lVert X-Y_n\rVert_{2}^{2}\ge0$. The display shows $d_m\ge d_n$ for $m\le n$, so the set $\{d_n:n\in\mathbb{N}\}$ is bounded below by $0$ and has a greatest lower bound $L\ge0$, which exists by the least upper bound property applied to the set of lower bounds. Given a real $\varepsilon>0$, since $L+\varepsilon^{2}$ is not a lower bound there is $N\in\mathbb{N}$ with $d_N<L+\varepsilon^{2}$. For $n,m\ge N$ with, say, $m\le n$, monotonicity of $(d_n)$ and the display give

Hence $(Y_n)_{n\in\mathbb{N}}$ is Cauchy in mean square.

Step 2: A mean-square limit measurable with respect to the limit information. Each $Y_n$ is $\mathcal{G}_n$-measurable and $\mathcal{G}_n\subseteq\mathcal{G}_\infty$, so each $Y_n$ is $\mathcal{G}_\infty$-measurable. By mean-square completeness relative to a sub-$\sigma$-algebra, applied with the sub-$\sigma$-algebra $\mathcal{G}_\infty$, there exists a $\mathcal{G}_\infty$-measurable square-integrable random variable $Y'$ with $\lVert Y_n-Y'\rVert_{2}\to0$.

Step 3: $Y'$ is a conditional expectation of $X$ given $\mathcal{G}_\infty$. It remains to verify the averaging property (iii) of Conditional Expectation of a Square-Integrable Random Variable. For an event $A\in\mathcal{F}$ the indicator $\mathbf{1}_{A}$ is a nonnegative simple function, hence a square-integrable random variable with $\lVert\mathbf{1}_{A}\rVert_{2}\le1$.

First let $A\in\bigcup_{n}\mathcal{G}_n$, say $A\in\mathcal{G}_m$. For every $n\ge m$ we have $A\in\mathcal{G}_n$, so the averaging property of $Y_n$ gives $\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[Y_n\mathbf{1}_{A}]$. By the Cauchy-Schwarz inequality,

using the linearity of expectation from Linearity and Monotonicity of the Lebesgue Integral. Hence $\mathbb{E}[Y'\mathbf{1}_{A}]=\mathbb{E}[X\mathbf{1}_{A}]$ for every $A\in\bigcup_n\mathcal{G}_n$.

Now extend to $\mathcal{G}_\infty$ by Dynkin's $\pi$-$\lambda$ theorem. The family $\mathcal{P}=\bigcup_n\mathcal{G}_n$ is a $\pi$-system: it is nonempty ($\Omega\in\mathcal{G}_1$), and if $A\in\mathcal{G}_j$ and $B\in\mathcal{G}_k$ then both lie in $\mathcal{G}_{\max(j,k)}$ by the nondecreasing hypothesis, so $A\cap B\in\mathcal{G}_{\max(j,k)}\subseteq\mathcal{P}$. Let

We check that $\mathcal{L}$ is a $\lambda$-system in the sense of Dynkin's Pi-Lambda Theorem. (1) $\Omega\in\mathcal{P}\subseteq\mathcal{L}$. (2) If $A,B\in\mathcal{L}$ with $A\subseteq B$, then pointwise $\mathbf{1}_{B\setminus A}=\mathbf{1}_{B}-\mathbf{1}_{A}$, and the linearity of expectation from Linearity and Monotonicity of the Lebesgue Integral (applied to the integrable products, which are integrable by Square-Integrable Random Variables and the Mean-Square Inner Product) gives $\mathbb{E}[Y'\mathbf{1}_{B\setminus A}]=\mathbb{E}[Y'\mathbf{1}_{B}]-\mathbb{E}[Y'\mathbf{1}_{A}]=\mathbb{E}[X\mathbf{1}_{B}]-\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[X\mathbf{1}_{B\setminus A}]$, so $B\setminus A\in\mathcal{L}$. (3) Let $(A_k)_{k\in\mathbb{N}}$ be a nondecreasing sequence in $\mathcal{L}$ with union $A$; note $A\in\mathcal{G}_\infty$ since $\mathcal{G}_\infty$ is a $\sigma$-algebra. The sets $A_1$ and $A_{k+1}\setminus A_k$ ($k\in\mathbb{N}$) are pairwise disjoint with union $A$, so countable additivity of the probability measure $P$ gives $P(A)=P(A_1)+\sum_{k}P(A_{k+1}\setminus A_k)$, and the partial sums telescope to $P(A_k)$; hence $P(A\setminus A_k)=P(A)-P(A_k)\to0$. Since pointwise $\mathbf{1}_{A}-\mathbf{1}_{A_k}=\mathbf{1}_{A\setminus A_k}$ and $\lVert\mathbf{1}_{A\setminus A_k}\rVert_{2}^{2}=P(A\setminus A_k)$, the Cauchy-Schwarz inequality yields

and the same bound with $X$ in place of $Y'$. Passing to the limit in $\mathbb{E}[Y'\mathbf{1}_{A_k}]=\mathbb{E}[X\mathbf{1}_{A_k}]$ gives $A\in\mathcal{L}$. By Dynkin's Pi-Lambda Theorem, $\mathcal{G}_\infty=\sigma(\mathcal{P})\subseteq\mathcal{L}$. Together with Step 2, $Y'$ satisfies (i)-(iii) of Conditional Expectation of a Square-Integrable Random Variable for the sub-$\sigma$-algebra $\mathcal{G}_\infty$, so $Y'$ is a conditional expectation of $X$ given $\mathcal{G}_\infty$.

Step 4: Conclusion. By the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, $P(Y_\infty=Y')=1$, and then $\lVert Y'-Y_\infty\rVert_{2}=0$ by the null-equivalence statement of Square-Integrable Random Variables and the Mean-Square Inner Product. The triangle inequality gives

which is the asserted limit. $\blacksquare$