Let $(E,\mathcal{A},\mu)$ be probability space and $\{f\_n\}$ be sequence of functions such that
$$
\sup\_n\int\_{E}|f\_n|d\mu<+\infty.
$$
Let $\{B\_p\}$ be a sequence non-increasing in $\mathcal{A}$ such that $\mu(\cap\_p B\_p) =0$ and for every $p$
$$
\{f\_n\}\text{ is uniformly integrable over }E\setminus B\_p
$$
Can we say that $\{f\_n\}\text{ is uniformly integrable over }E\setminus (\cap\_p B\_p)$?



Can we say that $\{f_n\}\text{ is uniformly integrable over }E\setminus (\cap_p B_p)$?
No.


Take $E=[0,1]$ with Lebesgue measure. Let $f\_n = n 1\_{[0, 1/n]}$, so that $\int\_E |f\_n|\,d\mu = 1$ for every $n$, and $B\_p = [0, 1/p]$. Note that $|f\_n| \le p$ on $E \setminus B\_p = (1/p, 1]$ for every $n$, so that $\{f\_n\}$ is indeed uniformly integrable over $E \setminus B\_p$. But clearly $\{f\_n\}$ is not uniformly integrable over $E \setminus \bigcap\_p B\_p = (0,1]$.


