# Is this family of functions uniformly integrable over $[0,1]$

Let $\mathcal F$ be a family of functions on $[0,1]$ each of which is integrable over $[0,1]$ and has $\int_a^b|f|\le b-a$ for all $[a,b] \subseteq [0,1]$. Is $\mathcal F$ uniformly integrable over $[0,1]$?

I think it is U.I. and this is what I have done so far. I wanted to see if this was right.

Let $\epsilon >0$ and let $\delta = \epsilon$. For any measurable set $A \subseteq [0,1]$ with $m(A) < \epsilon$ there exists a finite collection of open intervals, $\{I_k\}_{k=1}^n$, such that $m(I_k) < \frac{\epsilon}{n}$ for each $k$ and $A \subset \cup_{k=1}^nI_k$. This finite covering exists because any bounded subset of $\Bbb R$ is totally bounded. Then since $\int_A |f| \le \int_{\cup_{k=1}^nI_k}|f| \le \sum_{k=1}^n \int_{I_k}|f| \le \sum_{k=1}^n m(I_k) < \sum_{k=1}^n\frac{\epsilon}{n}=\epsilon$ then this implies that $\mathcal F$ is uniformly integrable over $[0,1]$. Is this right?

• The fact that $A$ is totally bounded says $A$ can be covered by finitely many intervals of length $\epsilon/n$. It does not say that $A$ can be covered by $n$ intervals of length $\epsilon/n$. If that were right then every subset of $[0,1]$ would have measure zero. But $m(A)<\epsilon$ does say that $A$ can be covered by countably many intervals $I_k$ with $\sum m(I_k)<\epsilon$. Use that. – David C. Ullrich Jun 8 '16 at 20:42

1. If $O$ is an open subset of $[0,1]$, then $O$ can be written as a countable disjoint union of open intervals.
2. For each Borel-measurable subset $B$ of $[0,1]$ and each positive $\delta$, there exists a closed set $F$ and an open set $O$ such that $F\subset B\subset O$ and $\lambda\left(O\setminus F\right)\lt\delta$.
Use fact 1. in order to prove that for each open set $O$, and each $f\in\mathcal F$, $\int_O\left|f\right|\mathrm d\lambda\leqslant \lambda\left(O\right)$. Then use fact 2. to control $\int_B\left|f\right|\mathrm d\lambda$.