A uniformly integrable family need not admit a single integrable majorant

The series ${\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k^2}$ converges to a value below $1$, so the intervals $I_k$ are pairwise disjoint subsets of $[0,1]$. 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 ${\int }_{\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}}\mathrm{\mid }{f}_k\mathrm{\mid }d\lambda =\frac{1}{2k}\le \frac{1}{2\lfloor M\rfloor +2}\mathrm{.}$ Hence the family is uniformly integrable by [L1].

If $g$ were an integrable majorant for all $f_k$, then $\mathrm{\mid }g\mathrm{\mid }\ge g\ge k$ almost everywhere on $I_k$. Since the intervals are pairwise disjoint, $\int \mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}{\int }_{I_k}\mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k}=+\mathrm{\infty },$ contradicting integrability of $g$.

This family is therefore uniformly integrable but has no single integrable majorant. The example is exactly the failure of the false statement paired with [L2].