FALSE: uniform integrability implies domination by one integrable function

The intervals $I_k$ are pairwise disjoint and lie in $[0,1]$ because ${\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k^2}<1$. Also $\int \mathrm{\mid }{f}_k\mathrm{\mid }d\lambda =k\lambda (I_k)=\frac{1}{2k}\mathrm{.}$

If $M>0$ and $k\le M$, then $\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}=\mathrm{\varnothing }$; if $k>M$, then $\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}=I_k$ and ${\int }_{\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}}\mathrm{\mid }{f}_k\mathrm{\mid }d\lambda =\frac{1}{2k}\le \frac{1}{2\lfloor M\rfloor +2}\mathrm{.}$ Therefore the family $(f_k)$ is uniformly integrable by [L1].

If an integrable function $g$ satisfied $g\ge f_k$ almost everywhere for every $k$, then $\mathrm{\mid }g\mathrm{\mid }\ge g\ge k$ almost everywhere on $I_k$. Since the intervals are pairwise disjoint, $\int \mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}{\int }_{I_k}\mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}k\lambda (I_k)={\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k}=+\mathrm{\infty },$ contradicting integrability of $g$. Hence no single integrable majorant exists.

This uniformly integrable family is not dominated by any integrable function, so the claim is false.