Can we say that $\{f_n\}\text{ is uniformly integrable over }E\setminus (\cap_p B_p)$? [closed]

Question

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 $\{f_n\}\text{ is uniformly integrable over }E\setminus (\cap_p B_p)$?

Answer 1

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 $E \setminus \bigcap_p B_p = (0,1]$.