Proof of Generated Sigma-Algebras Need Not Converge in Mean Square under Convergence of the Generating Random Variables

Construction. Let $\Omega=\{0,1\}\times\{0,1\}$ be the Cartesian product of two copies of $\{0,1\}$, let $\mathcal{F}$ be the family of all subsets of $\Omega$, which is a $\sigma$-algebra, and let $P(A)$ be the number of elements of $A$ divided by $4$. Then $P$ is a probability measure: $P(\emptyset)=0$, $P(\Omega)=1$, and any countable disjoint union of subsets of the four-element set $\Omega$ has only finitely many nonempty terms, so countable additivity reduces to finite additivity of counting. Thus $(\Omega,\mathcal{F},P)$ is a probability space. Define, for $\omega=(\omega_1,\omega_2)\in\Omega$,

$$U(\omega)=\omega_1,\qquad V(\omega)=\omega_2,\qquad U^n(\omega)=U(\omega)+\tfrac{1}{2n}\,V(\omega)\quad(n\in\mathbb{N}).$$

Since $\mathcal{F}$ contains every subset of $\Omega$, every real-valued function on $\Omega$ is a random variable. Each of $U$, $V$, $U^n$ takes values in $[0,2]$, so its square is bounded by $4$ pointwise and has finite expectation by the monotonicity of the integral in Linearity and Monotonicity of the Lebesgue Integral; hence all are square-integrable.

Part (a). For every $\omega\in\Omega$ and $n\in\mathbb{N}$, $|U^n(\omega)-U(\omega)|=\frac{1}{2n}V(\omega)\le\frac{1}{2n}$, so $U^n(\omega)\to U(\omega)$ for every $\omega$. Moreover $(U^n-U)^{2}\le\frac{1}{4n^{2}}$ pointwise, so by monotonicity of the integral from Linearity and Monotonicity of the Lebesgue Integral, $\lVert U^n-U\rVert_{2}\le\frac{1}{2n}$, and this real sequence has limit $0$.

Part (b). Fix $n\in\mathbb{N}$. The four points $(0,0),(0,1),(1,0),(1,1)$ of $\Omega$ are mapped by $U^n$ to the four values $0,\ \frac{1}{2n},\ 1,\ 1+\frac{1}{2n}$, which are pairwise distinct because $0<\frac{1}{2n}\le\frac12<1$. For any real $c$, the singleton $\{c\}$ is a Borel set, since its complement $(-\infty,c)\cup(c,\infty)$ is open. Hence for each $\omega\in\Omega$ the preimage $(U^n)^{-1}(\{U^n(\omega)\})=\{\omega\}$ belongs to the generated $\sigma$-algebra $\sigma(U^n)$. Every subset of $\Omega$ is a finite union of singletons, so $\sigma(U^n)=\mathcal{F}$. In particular $V^{-1}(B)\in\mathcal{F}=\sigma(U^n)$ for every Borel set $B$, i.e., $V$ is $\sigma(U^n)$-measurable.

Part (c). By Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, $\sigma(U)=\{U^{-1}(B):B\ \text{Borel}\}$. For any Borel set $B$, the preimage $U^{-1}(B)$ depends only on which of $0,1$ lie in $B$, so with $A_0=\{0\}\times\{0,1\}$ and $A_1=\{1\}\times\{0,1\}$ we get $\sigma(U)=\{\emptyset,\ A_0,\ A_1,\ \Omega\}$. The set $V^{-1}(\{1\})=\{(0,1),(1,1)\}$ is nonempty, omits $(0,0)$ so is not $\Omega$, contains $(1,1)\notin A_0$ so is not $A_0$, and contains $(0,1)\notin A_1$ so is not $A_1$; hence $V^{-1}(\{1\})\notin\sigma(U)$, and since $\{1\}$ is Borel, $V$ is not $\sigma(U)$-measurable. The constant sequence $X^n=V$ is $\sigma(U^n)$-measurable for every $n$ by Part (b), square-integrable, and satisfies $\lVert X^n-V\rVert_{2}=0$ for all $n$, so it converges in mean square to $V$, which is not $\sigma(U)$-measurable.

Part (d). Since $V$ is $\sigma(U^n)$-measurable and square-integrable, $V$ itself satisfies conditions (i)-(iii) of Conditional Expectation of a Square-Integrable Random Variable for the sub-$\sigma$-algebra $\sigma(U^n)$, and by the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables every conditional expectation of $V$ given $\sigma(U^n)$ equals $V$ almost surely.

Next, $\sigma(V)=\{\emptyset,B_0,B_1,\Omega\}$ with $B_k=\{0,1\}\times\{k\}$, by the argument of Part (c) applied to $V$. The $\sigma$-algebras $\sigma(V)$ and $\sigma(U)$ are independent: for the nontrivial pairs, $P(A_j\cap B_k)=P(\{(j,k)\})=\tfrac14=P(A_j)\,P(B_k)$ since $P(A_j)=P(B_k)=\tfrac12$, and pairs involving $\emptyset$ or $\Omega$ are immediate. The expectation of $V$ is computed from $V=\mathbf{1}_{B_1}$ as a simple function: $\mathbb{E}[V]=P(B_1)=\tfrac12$. By the independence property (property 5 of Basic Properties of Conditional Expectation for Square-Integrable Random Variables), the constant random variable $\tfrac12$ is a conditional expectation of $V$ given $\sigma(U)$, and every conditional expectation of $V$ given $\sigma(U)$ equals $\tfrac12$ almost surely by uniqueness.

Now fix $n$ and let $Y_n$ be any conditional expectation of $V$ given $\sigma(U^n)$ and $Y$ any conditional expectation of $V$ given $\sigma(U)$. Then $P(Y_n-Y=V-\tfrac12)=1$, so $\lVert (Y_n-Y)-(V-\tfrac12)\rVert_{2}=0$ by the null-equivalence statement of Square-Integrable Random Variables and the Mean-Square Inner Product, and the triangle inequality gives $\lVert Y_n-Y\rVert_{2}=\lVert V-\tfrac12\rVert_{2}$. Pointwise $(V-\tfrac12)^{2}=\tfrac14$ because $V$ takes only the values $0$ and $1$; hence $\lVert V-\tfrac12\rVert_{2}^{2}=\tfrac14$ and $\lVert Y_n-Y\rVert_{2}=\tfrac12$ for every $n$, proving the displayed identity in the statement. The constant sequence $\tfrac12$ does not have limit $0$, so applying Mean-Square Convergence of Sub-Sigma-Algebras with the square-integrable random variable $X=V$ shows that $\sigma(U^n)$ does not converge in mean square to $\sigma(U)$.

The final consequence recorded in the statement combines Parts (a), (c), and (d). $\blacksquare$