= Solution
A family $(Y_i)_{i\in I}$ of integrable <random variables> is <uniformly integrable> if
$$
\boxed{\lim_{K\to\infty}\sup_{i\in I}\mathbb E\bigl[|Y_i|\mathbf1_{\{|Y_i|>K\}}\bigr]=0.}
$$
This 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 $|X_n|\leq\mathbb E[|Z|\mid\mathcal F_n]$, hence $\mathbb E|X_n|\leq\mathbb E|Z|$. The <Markov inequality> therefore gives
$$
\mathbb P(A_n(K))\leq\frac{\mathbb E|Z|}{K}.
$$
Using the measurability of $A_n(K)$ in the defining integral identity for <conditional expectation>, for every $R>0$ we get
$$
\begin{aligned}
\mathbb E[|X_n|\mathbf1_{A_n(K)}]
&\leq\mathbb E[|Z|\mathbf1_{A_n(K)}]\\
&\leq\mathbb E[|Z|\mathbf1_{\{|Z|>R\}}]+R\mathbb P(A_n(K))\\
&\leq\mathbb E[|Z|\mathbf1_{\{|Z|>R\}}]+\frac{R\mathbb E|Z|}{K}.
\end{aligned}
$$
The first term tends to zero as $R\to\infty$ because $Z$ is integrable, by the <dominated convergence theorem>. Given $\varepsilon>0$, first choose $R$ to make it less than $\varepsilon/2$, then choose $K$ to make the second term less than $\varepsilon/2$. These choices work for every $n$. This proves <uniform integrability of conditional expectations>:
$$
\boxed{\{\mathbb E[Z\mid\mathcal F_n]:n\geq1\}\text{ is uniformly integrable}.}
$$
In fact the argument works for an arbitrary family of <sigma-algebras>; nesting is not needed for this conclusion.
Back to article page
