Given $\lim_{n\to \infty}\int_A(f_{n+1}-f_n)^2d\mu=0$ ,$ \int_Xf_n^2d\mu \le k $, is {$f_n $} uniformly integrable on $A$?

Question

Given $\lim_{n\to \infty}\int_A(f_{n+1}-f_n)^2d\mu=0$ , is {$f_n $} uniformly integrable on $A$ ?

$ f_n: R \to R $

$ n \in N $

$\mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. $ \int_Xf_n^2d\mu \le k $ where $ k \in R $ , $ X \subseteq A $

Answer 1

The question keeps changing, and the quantifiers are still a bit unclear as of now. If we're assuming that $\int_Xf_n^2\le k$ for every $X\subset A$ then the answer is yes, even without assuming that $\int_A(f_n-f_{n+1})^2\to0$.

Because then $\int_Af_n^2\le k$, so Cauchy-Schwarz shows that $$\int_E|f_n|\le\mu(E)^{1/2}k^{1/2}$$for every $E\subset A$.