= Solution
A family $\mathcal R$ is <uniformly integrable> if
$$
\lim_{K\to\infty}\sup_{X\in\mathcal R}E[|X|\mathbf1_{\{|X|>K\}}]=0.
$$
This implies integrability and a uniform bound on $E|X|$, since $E|X|\leq K+E[|X|\mathbf1_{\{|X|>K\}}]$. <Convergence in probability> means $P(|X_n-X|>\varepsilon)\to0$ for every $\varepsilon>0$; convergence in $L^1$ means $E|X_n-X|\to0$. 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 bound
$$
|X+Y|\mathbf1_{\{|X+Y|>K\}}
\leq2|X|\mathbf1_{\{|X|>K/2\}}+2|Y|\mathbf1_{\{|Y|>K/2\}}.
$$
Indeed, 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 \b[the whole family of sums is <uniformly integrable>], with no <independence> assumption.
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