# $f_n$ uniformly absolutely continuous implies $f_n^+$ uniformly absolutely continuous?

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!

• somewhere in there you forgot to say "for all $n$" – zhw. Oct 11 '17 at 22:07
• @zhw. I did’t forget, I asumed it was clear from ‘uniform’. Anyway I’ll edit the question to avoid any possible misinterpretation. – Jonatan B. Bastos Oct 11 '17 at 23:35

Hint: $\int_A f^+\, d\mu = \int_{A\cap \{f\ge0\}} f\, d\mu.$
• I'm assuming that you mean $f_n$ in the RHS. This was my first approach, but the difficulty here is that you need to relate that to $\left|\int_Af_n\,d\mu\right|$, and there is no obvious way to do that as the latter could be small and still the former be big. Unless I am missing something... (?) – Jonatan B. Bastos Oct 12 '17 at 21:36
• @JonatanB.Bastos Yes there was a typo there. Now edited. Note that $\mu({A\cap \{f_n\ge0\}}) \le \mu (A).$ – zhw. Oct 15 '17 at 21:45