 Solution

A family $(Y_i{)}_{i\in I}$ of integrable random variables is uniformly integrable ifThis controls the contributions of large values uniformly over the family, rather than merely bounding its L1 norms.

Put $X_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ and $A_n(K)=\{|X_n|>K\}\in {\mathcal{F}}_n$. The conditional Jensen inequality gives $|X_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]$, hence $\mathbb{E}|X_n|\le \mathbb{E}|Z|$. The Markov inequality therefore givesUsing the measurability of $A_n(K)$ in the defining integral identity for conditional expectation, for every $R>0$ we getThe first term tends to zero as $R\to \infty$ because $Z$ is integrable, by the dominated convergence theorem. Given $\epsilon >0$, first choose $R$ to make it less than $\epsilon /2$, then choose $K$ to make the second term less than $\epsilon /2$. These choices work for every $n$. This proves uniform integrability of conditional expectations:In fact the argument works for an arbitrary family of sigma-algebras; nesting is not needed for this conclusion.