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I want to prove the following statement:

If $f$ has uniformly absolutely continuous integral, then $|f|$ also has uniformly absolutely continuous integral.

This is a technical lemma I need to pass before I can prove Vitali's theorem. The rest is already proved elsewhere in the forum. Here we have a finite positive measure on a measure space $X$. Now knowing for any $\epsilon$, there is $\delta$ such that $\mu(E)<\delta$ implies $$|\int_{E}f d\mu|<\epsilon$$I need to show I can strength this to the uniformly absolutely continuous integral quality for $|f|$ as well. But I do not know how. If $X$ is a compact metric space, then a covering lemma would suffice. But in general there is no reason to assume $X$ is either compact or metrizable. So even though I can approximate $f$ on $X$ by a continuous function, $f$ might not be uniformly continuous.

In general case if $\frac{\int_{E} fd\mu}{\mu(E)}\in S$ for some closed set $S$, then I can conclude that $f$ must have value in $S$ almost everywhere. In this case, however we only know $\mu(E)<\delta$, so the closed set $S$ become $[\frac{\epsilon}{\delta},\infty]$ and is essentially useless. In theory $f$ could have a very strong oscillating quality that made $|f|$'s integral problematic, and I do not know how to rule out that.

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Bombyx mori
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You can work by contradiction. The idea is to show that if $f$ oscillates wildly enough to screw with the integrability of $|f|$, then $f$ is not integrable on the support of all the positive peaks. Denote the positive and negative parts of $f$ by $f^+$ and $f^-$, respectively.

Suppose there is some $\epsilon$ such that for any $\delta$, there exists a set $E$ such that $\mu(E) < \delta$ but $\int_E|f| d\mu > \epsilon$. Then either $\int_E f^+ d\mu \geq \epsilon/2$ or $\int_E f^- d\mu \geq \epsilon/2$. We assume the former without loss of generality.

Now consider the (measurable) set $E^+ = E \cap \{f \geq 0\}$. We know that $\int_{E^+} f d\mu = \int_{E^+} f^+ d\mu > \frac{\epsilon}{2}$ despite the fact that $\mu(E^+) \leq \mu(E) < \delta$. We can find such a set for an $\delta$. This implies that the integral of $f$ is not uniformly absolutely continuous.

(It never ceases to amaze me that the $f$ has so much control over $|f|$!)

Adam Saltz
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  • This is essentially a rephrasing of a standard proof about absolutely continuous measures, right? – Potato Jan 03 '13 at 06:13
  • @Potato Something about it feels familiar, but I don't have enough experience to say which proofs (or facts) are standard. – Adam Saltz Jan 03 '13 at 06:15
  • There is a technical problem in your proof; you want $\int_{E^{+}}fd\mu>\epsilon$,but if I am not mistaken with your assumption you can only have it $>\epsilon/2$. In fact if $\int_{E}|f|d\mu>\epsilon$, then either $\int_{E}f^{+}d\mu$ or $\int_{E}f^{-}d\mu$ must be greater than or equal to $\epsilon$. This should not be difficult. – Bombyx mori Jan 03 '13 at 07:33
  • @user32240 I have replaced the last $\epsilon$ with $\epsilon/2$. Thanks.

    One could certain go back and start with $2\epsilon$, but I've never seen much benefit in this. Moreover, I think it makes a proof much more natural to read without unmotivated constant factors floating around. Anyone familiar with $\delta$-$\epsilon$ judo (like you!) should be able to adjust the proof to satisfaction.

     – Adam Saltz Jan 03 '13 at 15:10