[proofplan] We use the definition of [uniform integrability](/page/Uniform%20Integrability) by two uniform estimates: boundedness in $L^1(E)$ and absolute continuity of the integrals over small measurable sets. For the linear-combination family, the pointwise triangle inequality bounds $|af+bg|$ by $|a||f|+|b||g|$, so both estimates split into the corresponding estimates for $\mathcal F$ and $\mathcal G$. For the dominated family $\mathcal H$, each $h\in\mathcal H$ is controlled almost everywhere by some $f\in\mathcal F$, and monotonicity of the integral transfers both uniform estimates from $\mathcal F$ to $\mathcal H$. [/proofplan] [step:Record the uniform integrability estimates for the two original families] We use the standard definition of uniform integrability: a family $\mathcal A\subset L^1(E,\mathcal E,\mu)$ is uniformly integrable if \begin{align*} \sup_{u\in\mathcal A}\int_E |u|\,d\mu(x)<\infty \end{align*} and, for every $\varepsilon>0$, there exists $\delta>0$ such that every $A\in\mathcal E$ with $\mu(A)<\delta$ satisfies \begin{align*} \sup_{u\in\mathcal A}\int_A |u|\,d\mu(x)<\varepsilon. \end{align*} Here the supremum over an empty indexing family is interpreted as $0$. Since $\mathcal F$ and $\mathcal G$ are uniformly integrable, define \begin{align*} M_{\mathcal F}:=\sup_{f\in\mathcal F}\int_E |f|\,d\mu(x) \end{align*} and \begin{align*} M_{\mathcal G}:=\sup_{g\in\mathcal G}\int_E |g|\,d\mu(x). \end{align*} Then $M_{\mathcal F}<\infty$ and $M_{\mathcal G}<\infty$. Also, for each $\eta>0$, there are positive numbers $\delta_{\mathcal F}(\eta)>0$ and $\delta_{\mathcal G}(\eta)>0$ such that, whenever $A\in\mathcal E$, \begin{align*} \mu(A)<\delta_{\mathcal F}(\eta)\implies \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\eta \end{align*} and \begin{align*} \mu(A)<\delta_{\mathcal G}(\eta)\implies \sup_{g\in\mathcal G}\int_A |g|\,d\mu(x)<\eta. \end{align*} [/step] [step:Prove uniform integrability of fixed linear combinations] Let $\alpha:=|a|$ and $\beta:=|b|$. For arbitrary $f\in\mathcal F$ and $g\in\mathcal G$, the representative $af+bg$ is integrable because \begin{align*} |af+bg|\le \alpha |f|+\beta |g| \end{align*} $\mu$-a.e. Hence \begin{align*} \int_E |af+bg|\,d\mu(x)\le \alpha\int_E |f|\,d\mu(x)+\beta\int_E |g|\,d\mu(x). \end{align*} Taking the supremum over $f\in\mathcal F$ and $g\in\mathcal G$ gives \begin{align*} \sup_{u\in a\mathcal F+b\mathcal G}\int_E |u|\,d\mu(x)\le \alpha M_{\mathcal F}+\beta M_{\mathcal G}<\infty. \end{align*} It remains to prove the small-set estimate. Let $\varepsilon>0$ be given. If $\alpha>0$, set \begin{align*} \eta_{\mathcal F}:=\frac{\varepsilon}{2\alpha}. \end{align*} If $\alpha=0$, choose any positive value for $\eta_{\mathcal F}$ and ignore the $\mathcal F$ contribution. If $\beta>0$, set \begin{align*} \eta_{\mathcal G}:=\frac{\varepsilon}{2\beta}. \end{align*} If $\beta=0$, choose any positive value for $\eta_{\mathcal G}$ and ignore the $\mathcal G$ contribution. Define \begin{align*} \delta:=\min\{\delta_{\mathcal F}(\eta_{\mathcal F}),\delta_{\mathcal G}(\eta_{\mathcal G})\}. \end{align*} For every $A\in\mathcal E$ with $\mu(A)<\delta$ and every $f\in\mathcal F$, $g\in\mathcal G$, the triangle inequality and monotonicity of the integral give \begin{align*} \int_A |af+bg|\,d\mu(x)\le \alpha\int_A |f|\,d\mu(x)+\beta\int_A |g|\,d\mu(x). \end{align*} If $\alpha>0$ and $\beta>0$, the definition of $\delta$ gives \begin{align*} \int_A |af+bg|\,d\mu(x)<\alpha\eta_{\mathcal F}+\beta\eta_{\mathcal G}=\varepsilon. \end{align*} If one of $\alpha,\beta$ is zero, the corresponding term is identically zero, and the same estimate follows from the nonzero term. If $\alpha=\beta=0$, the family $a\mathcal F+b\mathcal G$ consists only of the zero element of $L^1(E,\mathcal E,\mu)$ and the estimate is immediate. Therefore $a\mathcal F+b\mathcal G$ is uniformly integrable. [guided] Let us prove the two requirements in the definition of uniform integrability for the family $a\mathcal F+b\mathcal G$. First set $\alpha:=|a|$ and $\beta:=|b|$. For $f\in\mathcal F$ and $g\in\mathcal G$, the pointwise triangle inequality gives \begin{align*} |af+bg|\le |af|+|bg|=\alpha |f|+\beta |g| \end{align*} $\mu$-a.e. Since $f,g\in L^1(E,\mathcal E,\mu)$, the right-hand side is integrable, so $af+bg\in L^1(E,\mathcal E,\mu)$. Integrating the inequality over all of $E$ gives \begin{align*} \int_E |af+bg|\,d\mu(x)\le \alpha\int_E |f|\,d\mu(x)+\beta\int_E |g|\,d\mu(x). \end{align*} Now take the supremum over all choices of $f$ and $g$. The uniform $L^1$ bounds for $\mathcal F$ and $\mathcal G$ yield \begin{align*} \sup_{u\in a\mathcal F+b\mathcal G}\int_E |u|\,d\mu(x)\le \alpha M_{\mathcal F}+\beta M_{\mathcal G}<\infty. \end{align*} The second requirement is the absolute continuity of these integrals uniformly over the whole family. Fix $\varepsilon>0$. The only small issue is that $a$ or $b$ may be zero, so we split the error only among the nonzero coefficients. If $\alpha>0$, choose \begin{align*} \eta_{\mathcal F}:=\frac{\varepsilon}{2\alpha}. \end{align*} Then uniform integrability of $\mathcal F$ gives a number $\delta_{\mathcal F}(\eta_{\mathcal F})>0$ such that \begin{align*} \mu(A)<\delta_{\mathcal F}(\eta_{\mathcal F})\implies \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\eta_{\mathcal F}. \end{align*} If $\alpha=0$, the term involving $\mathcal F$ is multiplied by zero, so any positive value of $\eta_{\mathcal F}$ is harmless. We treat $\mathcal G$ in the same way: if $\beta>0$, set \begin{align*} \eta_{\mathcal G}:=\frac{\varepsilon}{2\beta}. \end{align*} If $\beta=0$, choose any positive value. Define \begin{align*} \delta:=\min\{\delta_{\mathcal F}(\eta_{\mathcal F}),\delta_{\mathcal G}(\eta_{\mathcal G})\}. \end{align*} Now let $A\in\mathcal E$ satisfy $\mu(A)<\delta$. For every $f\in\mathcal F$ and $g\in\mathcal G$, \begin{align*} \int_A |af+bg|\,d\mu(x)\le \alpha\int_A |f|\,d\mu(x)+\beta\int_A |g|\,d\mu(x). \end{align*} When both coefficients are nonzero, the two small-set estimates give \begin{align*} \int_A |af+bg|\,d\mu(x)<\alpha\frac{\varepsilon}{2\alpha}+\beta\frac{\varepsilon}{2\beta}=\varepsilon. \end{align*} When exactly one coefficient is zero, the vanished term contributes $0$ and the remaining estimate gives a bound strictly below $\varepsilon$. When both coefficients are zero, every element of $a\mathcal F+b\mathcal G$ is the zero $L^1$ class, so the integral over $A$ is $0$. Thus the small-set estimate holds uniformly over $a\mathcal F+b\mathcal G$, and this family is uniformly integrable. [/guided] [/step] [step:Transfer uniform integrability to the dominated family] We prove the two defining estimates for $\mathcal H$. For each $h\in\mathcal H$, choose $f_h\in\mathcal F$ such that \begin{align*} |h|\le |f_h| \end{align*} $\mu$-a.e. By monotonicity of the integral, \begin{align*} \int_E |h|\,d\mu(x)\le \int_E |f_h|\,d\mu(x)\le M_{\mathcal F}. \end{align*} Taking the supremum over $h\in\mathcal H$ gives \begin{align*} \sup_{h\in\mathcal H}\int_E |h|\,d\mu(x)\le M_{\mathcal F}<\infty. \end{align*} Let $\varepsilon>0$. By uniform integrability of $\mathcal F$, choose $\delta>0$ such that every $A\in\mathcal E$ with $\mu(A)<\delta$ satisfies \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\varepsilon. \end{align*} For such an $A$ and for any $h\in\mathcal H$, the chosen $f_h\in\mathcal F$ satisfies \begin{align*} \int_A |h|\,d\mu(x)\le \int_A |f_h|\,d\mu(x)<\varepsilon. \end{align*} Taking the supremum over $h\in\mathcal H$ gives \begin{align*} \sup_{h\in\mathcal H}\int_A |h|\,d\mu(x)<\varepsilon. \end{align*} Thus $\mathcal H$ is uniformly integrable. This completes the proof of both assertions. [/step]