Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 25J / Solution

A family $\mathcal{R}$ is uniformly integrable ifThis implies integrability and a uniform bound on $E|X|$, since $E|X|\le K+E[|X|1_{\{|X|>K\}}]$. Convergence in probability means $P(|X_n-X|>\epsilon )\to 0$ for every $\epsilon >0$; convergence in $L^1$ means $E|X_n-X|\to 0$. The latter implies the former by Markov's inequality. Conversely, convergence in probability together with uniform integrability of $\{X_n\}$ implies that $X$ is integrable and $X_n\to X$ in $L^1$. Moreover an $L^1$-convergent sequence is uniformly integrable. This is the criterion that excludes rare very large values from carrying a nonvanishing first-moment error.

For the sum family, use the pointwise boundIndeed, if the left indicator is one then the larger of $|X|,|Y|$ exceeds $K/2$, and twice that larger number bounds $|X+Y|$. Taking expectations and the supremum over ${\mathcal{R}}_1,{\mathcal{R}}_2$ bounds the sum-family tail by the sum of their two uniformly vanishing tails. Thus the whole family of sums is uniformly integrable, with no independence assumption.