Let $(X,\mathcal{A},\mu)$ be a measure space. Assume that $\mu$ is a finite measure and that $f_n$ is a uniformly absolutely continuous sequence; i.e., that given $\varepsilon>0$ there exists $\delta>0$ such that, if $A\in\mathcal{A}$ satisfies $\mu(A)<\delta$, then $$ \left|\int_Af_n\,d\mu\right|<\varepsilon $$ for all $n$. Does it follows that the sequence of positive parts $f_n^+$ is also uniformly absolutely continuous?
First I thought of using the triangular reversed inequality but that is of not much use since $\int_Af_n\,d\mu$ can be small and still $\int_Af_n^+\,d\mu$ be big. Also I tried to break $A$ into $A^+=A\cap\{f_n<0\}$ and $A^-:=A\cap\{f_n\geq0\}$ and use the uniform absolutely continuity of $f_n$, but that didn't help much because then $A^+$ and $A^-$ depend on $n$. Any idea on how to proceed? Thank you!
Hint: $\int_A f^+\, d\mu = \int_{A\cap \{f\ge0\}} f\, d\mu.$