Does almost sure convergence and $L^1$-convergence imply almost sure convergence of the conditional expectation?

Question

Question. Let $ X_{n}, X $ be random variables on some probability space $ ( \Omega, \mathcal{F},\mathbb{P} ) $ and let $ \mathcal{G} \subset \mathcal{F} $ be a sub-$\sigma$-algebra. Moreover suppose that $ X_{n } \to X $ a.s. and in $ \mathrm{L}^{ 1 } ( \Omega, \mathcal{F}, \mathbb{P}) $. Does this already imply that $ \mathbb{ E} ( X_{ n } | \mathcal{G} ) \to \mathbb{ E} ( X | \mathcal{G} ) $ a.s.?

We know that the $ \mathrm{L}^{1} $-convergence implies that $\mathbb{ E} ( X_{ n } | \mathcal{G} ) \to \mathbb{ E} ( X | \mathcal{G} ) $ in $\mathrm{L}^{ 1 } $. However, we can in general only expect a subsequence to converge a.s. This can be seen by looking at $\mathcal{ G } = \mathcal{ F } $ and taking the usual counterexample that $ \mathrm{L}^{ 1 } $ convergence does not always imply a.s. convergence. But what happens if we additionally assume the a.s. convergence of the original sequence?

Answer 1

No, in general this is not true.

The ingredients: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of non-negative random variables such that $Y_n \to 0$ in $L^1$ but $Y_n$ does not converge almost surely to $0$ (e.g. the classical example of a sequence converging in probability to $0$ but not almost surely; see Example 1 here). Let $(\xi_n)_{n \in \mathbb{N}}$ be a sequence of independent random variables such that $(Y_n)_{n \in \mathbb{N}}$ and $(\xi_n)_{n \in \mathbb{N}}$ are independent and $\mathbb{P}(\xi_n=2^n) = 2^{-n}$, $\mathbb{P}(\xi_n=0) = 1-2^{-n}$. If we set $$Z_n := \prod_{j=1}^n \xi_j,$$ then the Borel Cantelli lemma shows that $Z_n(\omega)= 0$ for $n\geq N(\omega)$ sufficiently large. Moreover, $\mathbb{E}(\xi_n)=1$ implies $\mathbb{E}(Z_n)=1$ for all $n \in \mathbb{N}$.

The counterexample: Define $X_n := Y_n \cdot Z_n$. Since $Z_n(\omega)=0$ for $n \geq N(\omega)$, we clearly have $X_n \to 0$ almost surely. Moreover, $X_n \geq 0$, and so

$$\mathbb{E}(|X_n|) = \mathbb{E}(X_n) = \mathbb{E}(Y_n) \underbrace{\mathbb{E}(Z_n)}_{=1} \to 0,$$

shows that $X_n \to 0$ in $L^1$. Hence, $X_n \to 0=:X$ almost surely in $L^1$. Now consider the sub $\sigma$-algebra $\mathcal{G} := \sigma(Y_n; n \geq 1)$. Then, by the independence,

$$\mathbb{E}(X_n \mid \mathcal{G}) = \mathbb{E}(Y_n Z_n \mid \mathcal{G}) = Y_n \mathbb{E}(Z_n \mid \mathcal{G}) = Y_n \underbrace{\mathbb{E}(Z_n)}_{=1}.$$

Since we have chosen $(Y_n)_{n \in \mathbb{N}}$ such that $Y_n$ does not converge almost surely to $0$, we conclude that $\mathbb{E}(X_n \mid \mathcal{G})=Y_n$ does not converge almost surely to $\mathbb{E}(X \mid \mathcal{G})=0$.