We defined a family of functions $\{f_i\}_{ i\in I}$ to be uniformly integrable if for every $\epsilon>0$ there is a number M s.t $\forall$ i:
$\int_{|f_i|>M}|f_i|dP<\epsilon$
a) $\{f_i\}_{i\in I}$ are uniformly integrable
b) $\sup_i\int|f_i|dP<\infty$.
1) Does $b) \Rightarrow a)$?
2) Does $a) \Rightarrow b)$?
So for 1) I think I have a good counter example to disprove, I used $ f_i=i\mathbb{1}_{[0,\frac{1}i]}$. Obviously $\sup_i\int|f_i|dP=1<\infty$. I want to prove that it's not U.I so by definition I want to show that there exists $\epsilon>0 \ s.t\ \forall M, \exists i\ s.t \int_{|f_i|>M}|f_i|dP\geq\epsilon$
So if we choose $\epsilon$ to be 1 then for every $i>M$ we will get $\int_{|f_i|>M} |f_i|dP=1$.
Now I'm not sure about proving $a) \Rightarrow b)$, what I did is the following:
Because $\{f_i\}_{ i\in I}$ are UI, $\exists M s.t\ \forall i\ \int_{|f_i|>M}|f_i|dP<3$, so we have:
$\sup_i\int|f_i|dP=\sup_i(\int_{|f_i|\leq M}|f_i|dP+\int_{|f_i|>M}|f_i|dP)\leq\sup_i\int_{|f_i|\leq M}MdP +3\leq \int_{0}^MMdP=M^2+3<\infty$
Is this correct?