Proof of Conditional Expectation and Estimation Error of the Controlled State

Claim 1. Each component family of $m^{\mathrm f}$ is mean-square continuous, since $m^{\mathrm f}$ is a mean-square solution by claim 2 of The Kalman-Bucy Filter Equation and Its Solution and mean-square continuity is part of that definition; each component family of $c$ is mean-square continuous by claim 1 of Superposition Decomposition of the Controlled State and Observations. Sums of mean-square continuous families are mean-square continuous (claim 1 of Basic Properties of the Mean-Square Riemann Integral), so each $(\widehat X^{i}_t)_t$ is mean-square continuous. By claim 3 of The Kalman-Bucy Filter Equation and Its Solution, each $(m^{\mathrm f}_t)^{i}$ is almost surely equal to a $\mathcal{G}_t$-measurable square-integrable random variable; by claim 3 of Superposition Decomposition of the Controlled State and Observations, so is each $c^{i}_t$; and finite linear combinations preserve this property by claim 2 of The Closed Mean-Square Span of a Family of Random Variables.

Claim 2. Fix $t$ and $i$. By claim 2 of Superposition Decomposition of the Controlled State and Observations, $X^{\alpha,i}_t=X^{i}_t+c^{i}_t$ almost surely. Since a conditional expectation of a square-integrable random variable given $\mathcal{G}_t$ depends on that random variable only through its almost sure class (the defining conditions involve only expectations, which are unchanged under almost sure modification), it suffices to identify the conditional expectations of $X^{i}_t+c^{i}_t$. By claim 1 of The Kalman-Bucy Filter Computes the Conditional Expectation in the Linear-Gaussian Model, some (hence, up to almost sure equality, every, by the uniqueness in Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables) conditional expectation $Y_1$ of $X^{i}_t$ given $\mathcal{G}_t$ satisfies $Y_1=(m^{\mathrm f}_t)^{i}$ almost surely. Let $\tilde c$ be a $\mathcal{G}_t$-measurable square-integrable version of $c^{i}_t$ (claim 3 of Superposition Decomposition of the Controlled State and Observations). Directly from Conditional Expectation of a Square-Integrable Random Variable, $\tilde c$ is a conditional expectation of itself given $\mathcal{G}_t$: it is $\mathcal{G}_t$-measurable, square-integrable, and the defining conditions hold trivially. By linearity (claim 1 of Basic Properties of Conditional Expectation for Square-Integrable Random Variables), $Y_1+\tilde c$ is a conditional expectation of $X^{i}_t+\tilde c$ given $\mathcal{G}_t$, hence of $X^{\alpha,i}_t$ (almost surely equal random variables). By the uniqueness in Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, every conditional expectation of $X^{\alpha,i}_t$ given $\mathcal{G}_t$ is almost surely equal to $Y_1+\tilde c=(m^{\mathrm f}_t)^{i}+c^{i}_t=\widehat X^{i}_t$ almost surely.

Claim 3. Almost surely, $X^{\alpha,i}_t-\widehat X^{i}_t=(X^{i}_t+c^{i}_t)-\bigl((m^{\mathrm f}_t)^{i}+c^{i}_t\bigr)=X^{i}_t-(m^{\mathrm f}_t)^{i}=e^{i}_t$, by claim 2 of Superposition Decomposition of the Controlled State and Observations. The remaining assertions are exactly claims 2 and 3 of The Kalman-Bucy Filter Computes the Conditional Expectation in the Linear-Gaussian Model.

Claim 4. Let $\tilde\zeta$ be a $\mathcal{G}_t$-measurable random variable almost surely equal to $\zeta$; it is square-integrable, since $\tilde\zeta=\zeta$ almost surely and second moments are unchanged under almost sure modification. Each $e^{i}_t$ is measurable with respect to $\sigma(e^{1}_t,\dots,e^{l}_t)$, since by Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras that $\sigma$-algebra contains the preimages of Borel sets under each component. By claim 3, $\sigma(e^{1}_t,\dots,e^{l}_t)$ and $\mathcal{G}_t$ are independent; since every preimage of a Borel set under $e^{i}_t$ lies in the former $\sigma$-algebra and every preimage under $\tilde\zeta$ lies in the latter, the random variables $e^{i}_t$ and $\tilde\zeta$ are independent. Both are square-integrable, hence integrable, so by Expectation of a Product of Independent Random Variables the product $e^{i}_t\tilde\zeta$ is integrable with $\mathbb{E}[e^{i}_t\tilde\zeta]=\mathbb{E}[e^{i}_t]\,\mathbb{E}[\tilde\zeta]=0$, since $\mathbb{E}[e^{i}_t]=0$ by claim 3. Finally $e^{i}_t\zeta=e^{i}_t\tilde\zeta$ almost surely, so $e^{i}_t\zeta$ is integrable with $\mathbb{E}[e^{i}_t\zeta]=0$. $\square$