convergence of conditional expectations

Question

Let $\xi$ be a measurable partition on a compact metric space $X$.

We can assume that $\xi$ is generated by a countable measurable sets $E_i$, that is, $\xi = \vee_{i=1}^{\infty} \{ E_i , X \backslash E_i \}$.

Now the conditional expectation $E(\phi | \mathcal{B}(\xi))$ of a bounded measurable function $\phi$ is defined by

$\lim_{n \rightarrow \infty } E(\phi | \mathcal{B}_n) $, where $\mathcal{B}_n$ is the finite sigma-algebra generated by sets $E_1,...,E_n$.

My question is:

if for a uniformly bounded measurable functions we have a pointwise limit $\phi_m \rightarrow \phi$, then $E(\phi_m | \mathcal{B}(\xi))$ converges to $E(\phi | \mathcal{B}(\xi))$ pointwise?

This statement is implicitly assumed in the proof of disintegration into conditional measures in http://w3.impa.br/~viana/out/rokhlin.pdf.

Answer 1

We forget the context of measurable partition of a compact metric space, we have the general result:

Proposition: Let $(\Omega,\mathcal F,\mu)$ be a probability space. If $\mathcal A\subset\mathcal F$ is a sub-$\sigma$-algebra $(X_n)_{n\geqslant 1}$ is a sequence of random variables converging pointwise to $X$ and such that $M:=\sup_{n,\omega}|X_n(\omega)|\lt \infty$, then $\mathbb E[X_n\mid\mathcal A]\to \mathbb E[X\mid\mathcal A]$ almost everywhere.

Proof: We use Egoroff's theorem: for a fixed $N$, consider $S_N\in\mathcal F$ such that $\sup_{\omega\in S_N}|X_j(\omega)|\to 0$ as $j\to \infty$ and $\mu(\Omega\setminus S_N)\leqslant 2^{-N}$. We have by Markov's inequality and Borel-Cantelli's lemma that $\mathbb E[\chi_{\Omega\setminus S_N}\mid\mathcal A]\to 0$ almost-everywhere as $N\to \infty$. We conclude using the inequality $$|\mathbb E[X_j\mid\mathcal A]|\leqslant \sup_{\omega\in S_N}|X_j(\omega)|+M\cdot\mathbb E[\chi_{\Omega\setminus S_N}\mid\mathcal A].$$

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Notice that the result is not true if we replace the assumption of uniform boundedness by uniform integrability.