TITLE: Uniform Integrability implies boundedness of $\sup_i\int|f_i|dP$?
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: For the second part in more general terms.
Note that if we assume $(f_i)_{i \in I}$ is uniformly integrable, then for all $M>0$ we have for all $i \in I$ that
$$
\int_\Omega |f_i| dP
= \int_{(|f_i| \leq M)} |f_i| dP + \int_{(|f_i| > M)} |f_i| dP \\
\leq M P(|f_i \leq M|) + \int_{(|f_i| > M)} |f_i| dP
\leq M + \int_{(|f_i| > M)} |f_i| dP.
$$
So also 
$$
\sup_{i \in I} \int |f_i| dP 
\leq M + \sup_{i \in I}\int_{(|f_i| > M)} |f_i| dP.
$$
The last term tends to zero as $M$ tends to infinity, so in particular for  $M$ sufficiently large we can conclude that $\sup_{i \in I} \int |f_i| dP$ is finite.
Would this be alright?