# An equivalent definition of uniform integrability

Let $(X,\mathcal{M},\mu)$ be a measure space and $\{f\}$ be a sequence of functions on $X$, each of which is integrable over $X$. Show that $\{f_n\}$ is uniformly integrable if and only if for each $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that for any natural number $n$ and measurable subset $E$ of $X$, if $\mu(E) \lt \delta$, $$\left|\int_E f_n~d\mu\right| \lt \epsilon.$$

I think one direction $(\Rightarrow)$ is clear. Since $f_n$ being uniformly integrable imply that for every $\varepsilon\gt 0,~\exists \delta \gt 0$ such that for any $E\in \mathcal{M},~\mu(E)\lt \delta$, $\int_E |f_n|~d\mu$ for every natural $n$. But then

$$\left|\int_E f_n~d\mu\right|\le \int_E|f_n|~d\mu \lt \varepsilon.$$

Any suggestions for the other direction?

Edit:

Following Davide's suggestions, I have

\begin{align*} \int_E |f_n|~d\mu & \le \int_{E\cap [f_n \ge 0]} f_n~d\mu + \int_{E\cap [f_n \lt 0]} (-f_n)~d\mu\\ & = \left| \int_{E\cap [f_n \ge 0]} f_n~d\mu \right| + \left| \int_{E\cap [f_n \lt 0]}f_n~d\mu \right| \\ & \lt \varepsilon + \varepsilon = 2\varepsilon. \end{align*}

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@DavideGiraudo: the definition I have is this: For every $\varepsilon \gt 0$, there is a $\delta\gt 0$ such that if $E$ is measurable and $\mu(E)\lt \delta$, then $\int_E |f_n|~d\mu \lt \varepsilon$ for all $n$. –  Kuku Mar 30 '12 at 11:29

1. For all $\varepsilon>0$, we can find $\delta >0$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$, then for all natural number $n$, we have $\int_E|f_n|d\mu\leq\varepsilon$;
2. For all $\varepsilon>0$, we can find $\delta >0$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$, then for all natural number $n$, we have $\left|\int_Ef_nd\mu\right|\leq\varepsilon$.
(1.) $\Rightarrow$ (2.) by the triangular inequality. To see the converse, fix $\varepsilon>0$. We can find $\delta$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$ then for each $n$: $\left|\int_Ef_nd\mu\right|\leq\varepsilon/2$. Let $E$ measurable with measure $\leq\delta$, and put $E_1:=\{x\in X:f(x)<0\}$ and $E_2:=\{x\in X:f(x)\geq 0\}$. These sets are measurable and of measure $\leq\delta$. We get for $n$ natural number: $$\int_E |f_n|d\mu=\int_{E_1} |f_n|d\mu+\int_{E_2} |f_n|d\mu=-\int_{E_1}f_nd\mu+ \int_{E_2}f_nd\mu\leq \left|\int_{E_1}f_nd\mu\right|+\left|\int_{E_2}f_nd\mu\right|\leq \varepsilon/2+\varepsilon/2=\varepsilon.$$