Exercise 1.5.13 (Equivalent formulation of uniform integrability I)

Question

Let $\left(fight)_{n\in\mathbb{N}}:X	o\mathbb{C}$ be a uniformly integrable sequence of functions. Show that for every $\epsilon>0$ there exists a $\delta>0$ such that for all $n\in\mathbb{N}$ and measurable sets $E \subset X$ with $\mu(E)\leq \delta$ we have $$\int_E |f_n| \ \mathrm{d}\mu \leq \epsilon.$$

Elu Thingol · Accepted Answer

By theorem assumption we already have some bounds where $|f_n|$ is than some very big bound or less than very small bound.
\begin{align*}
  \exists \overline{M}_\epsilon \in \mathbb{N} \quad \forall M\geq\overline{M}_\epsilon: \quad \sup_{n\in\mathbb{N}} \int_{\{|f_n|\geq M\}} |f_n| \leq \epsilon/3\
  \exists \underline{M}_\epsilon \in \mathbb{N} \quad \forall M\leq\underline{M}_\epsilon: \quad \sup_{n\in\mathbb{N}} \int_{\{|f_n|\leq M\}} |f_n| \leq \epsilon/3
\end{align*}
We see that if $\delta \leq \mu(\{|f_n|\geq \overline{M}\}), \mu(\{|f_n|\leq \underline{M}\})$, then the condition is satisfied for all sets $E \subseteq \{x \in X: |f_n|\leq \underline{M} 	ext{ or } |f_n|\geq \overline{M}\}$. However, what about the region in between $\{x \in X: \underline{M} \leq |f_n| \leq \overline{M}\}$? Here we have
$$\int_{\{\underline{M} \leq |f_n| \leq \overline{M}\}} |f_n| \leq \overline{M} \mu\left(\{x \in X: \underline{M} \leq |f_n| \leq \overline{M}\} ight).$$
In other words, for any set $F\subseteq \{x \in X: \underline{M} \leq |f_n| \leq \overline{M}\}$, the integral won't exceed $\overline{M}\cdot \mu(F)$. Thus, we set
$$\delta := \min\left\{ \begin{array}{c}
  \mu(\{|f_n|\geq \overline{M}_\epsilon\})\
  \mu(\{|f_n|\leq \underline{M}_\epsilon\}) \
     {\epsilon}/{3\overline{M}_\epsilon}
\end{array}
ight\}
$$
and achieve the theorem assertion in any region.

Exercise 1.5.13 (Equivalent formulation of uniform integrability I)

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Exercise 1.5.13 (Equivalent formulation of uniform integrability I)

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