Different forms of uniform integrability proof from Schilling

Question

This is part of a proof from Rene Schilling's Measures, Integrals, and Maringales. There is a part I don't understand here.

(vii). (a) $\sup_{u \in F} \int |u|d\mu <\infty$; (b) $\lim_{n\to \infty} \sup_{u\in F} \int_{A_n} u d\mu = 0$ for all decreasing sequences $(A_n) \subset \mathcal{A}$, $A_n \downarrow \emptyset$.

is equivalent to (viii). (a) $\sup_{u \in F} \int |u|d\mu <\infty$; (b) $\lim_{n\to \infty} \sup_{u\in F} \int_{A_n} u d\mu = 0$ for all decreasing sequences $(A_n) \subset \mathcal{A}$, $A_n \downarrow \emptyset$; (c) $\lim_{n\to \infty} \sup_{u\in F}\bigg| \int_{B_n} ud\mu\bigg|=0$ for all sequences $(B_n) \subset \mathcal{A}$ with $\mu(B_n)\to 0.$

Proof. (a) and (b) are the same. If (viii)(c) fails to hold, then there exists some $\epsilon>0$ such that $$\forall \delta>0 \;\exists B\in \mathcal{A},\mu(B)<\delta,u\in F:\bigg|\int_B ud\mu\bigg|\ge \epsilon. \; (22.6)$$

Set $\epsilon_k := \epsilon 4^{-k}.$ Recursively we construct sets $(B_n)\subset \mathcal{A}$ and $(u_n)\subset F$ such that

$$\max_{k\le n-1} \int_{B_n} |u_k|d\mu \le \epsilon_n \;\text{and}\;\bigg|\int_{B_n} u_n d\mu\bigg| \ge \epsilon.$$ For $n=1$, it is clear how to choose $B_1$ and $u_1$. Suppose that we have already constructed $B_1,\dots, B_{n-1}$ and $u_1,\dots ,u_{n-1};$ there exists some $\delta \in (0,1/n)$ such that $$\max_{k\le n-1} \int_B |u_k|d\mu \le \sum_{k=1}^{n-1}\int_B |u_k|d\mu \le \epsilon_n\; \text{for all}B\in \mathcal{A}, \mu(B)\le \delta.$$ Because of (22.6) there is some $B_n \in \mathcal{A}, \mu(B_n)\le \delta$, such that $\big| \int_{B_n} u_n d\mu\big| \ge \epsilon.$

I can't figure out the induction step. Using $B_1,\dots, B_{n-1}$ and $u_1,\dots ,u_{n-1},$ how do we get $\sum_{k=1}^{n-1}\int_B |u_k|d\mu \le \epsilon_n\; \text{for all}B\in \mathcal{A}, \mu(B)\le \delta$?

Answer 1

Recall the following statement:

For any function $f \in L^1(\mu)$ the following statement holds true: $$\forall \epsilon>0 \, \, \exists \delta>0 \, \, \forall B \in \mathcal{A},\mu(B)<\delta: \quad \int_B |f| \, d\mu \leq \epsilon.$$

By a very simple argument this result extends to any finite family $\{f_1,\ldots,f_k\} \subseteq L^1(\mu)$.

For any finite family $\{f_1,\ldots,f_k\} \subseteq L^1(\mu)$ the following statement holds true: $$\forall \epsilon>0 \, \, \exists \delta>0 \, \, \forall B \in \mathcal{A},\mu(B)<\delta: \quad \max_{j \leq k} \int_B |f_j| \, d\mu \leq \epsilon.$$

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Given the functions $u_1,\ldots,u_{n-1}$ (from the induction step) we can therefore choose $\delta>0$ such that

$$\int_B |u_j| \, d\mu \leq \epsilon_n$$

for all $j=1,\ldots,n-1$ and $B \in \mathcal{A}$ such that $\mu(B)<\delta$. Now (22.6) yields that there exists $B_n \in \mathcal{A}$, $\mu(B_n)<\delta$, and $u_n \in F$ such that $$\left| \int_{B_n} u_n \, d\mu \right| \geq \epsilon_n.$$