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How can I show that if family of $f$ is uniformly integrable then so is {$|f|$}?

$($by uniformly integrablity: $\forall \epsilon>0 \ \exists \delta>0: |\int_Ef|<\epsilon,\mu(E)<\delta)$

My attempt:

$|\int_E f|=|\int_{E^+} f^+-\int_{E^-} f^-| \leq \epsilon$

$|\int_{E^+} f^++\int_{E^-} f^--2\int_{E^-} f^-| \leq \epsilon$

$|\int_{E} |f|-2\int_{E^-} f^-| \leq \epsilon$

$\mu(E^-)\leq \mu(E)\leq \delta$ so I want to conclude $2\int_{E^-} f^-$ is negligible, thus $|\int_{E} |f|| \leq \epsilon$ and get done. Is it OK?

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Can you apply the property to a set $E \cap \{x| f(x) \geq 0 \}$ and $E \cap \{x| f(x) <0 \}$ separately and combine the results? – copper.hat Nov 5 '12 at 5:35

Let $\epsilon>0$, and find a $\delta>0$ such that if $\mu E<\delta$, then $|\int_E f| < \frac{\epsilon}{2}$.

Let $E$ be such that $\mu E < \delta$. Choose $f$ and let $P=\{x|f(x)\geq 0\}$, $N=\{x|f(x)<0\}$. Then $\mu (E \cap P) < \delta $, $\mu (E \cap N) < \delta $, and so $|\int_{E\cap P} f| < \frac{\epsilon}{2}$ and $|\int_{E \cap N} f| < \frac{\epsilon}{2}$. However, $\int_E |f| = \int_{E\cap P} f - \int_{E \cap N} f$, from which it follows that $\int_E |f| < \epsilon$.

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$\int_E |f| = \int_{E\cap P} f + \int_{E \cap N} f$ – Bunny Nov 5 '12 at 7:26
Nope, on $E \cap N$, we have $f(x) < 0$. So $\int_{E \cap N} |f| = - \int_{E \cap N} f$. (But, it doesn't matter either way.) – copper.hat Nov 5 '12 at 7:28
Sorry, I forgot you are using the $f$ itself not $f^+\ f^-$ – Bunny Nov 5 '12 at 7:31

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