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!

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

Hint: $\int_A f^+\, d\mu = \int_{A\cap \{f\ge0\}} f\, d\mu.$

  • $\begingroup$ 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... (?) $\endgroup$ – Jonatan B. Bastos Oct 12 '17 at 21:36
  • $\begingroup$ @JonatanB.Bastos Yes there was a typo there. Now edited. Note that $\mu({A\cap \{f_n\ge0\}}) \le \mu (A).$ $\endgroup$ – zhw. Oct 15 '17 at 21:45
  • $\begingroup$ Oh!! I don't know how come I didn't think about that... Thank you! $\endgroup$ – Jonatan B. Bastos Oct 16 '17 at 2:45

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