# Uniform Integrability implies boundedness of $\sup_i\int|f_i|dP$?

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?

• This seems correct, but not the part where you integrate $M$ from $0$ to $M$. We have $\int\vert f\vert I_{\vert f\vert\le M}\text{d}\mathbb{P}=\int_{\Omega}\vert f(\omega)\vert I_{\vert f\vert\le M}(\omega)\text{d}\mathbb{P}(\omega)\le \int_{\Omega} M\text{d}\mathbb{P}(\omega)= M\mathbb{P}(\Omega) = M$ – MoebiusCorzer Jan 20 '17 at 11:05
• Oh right because it's a probability measure, thank you for correcting me! – lfc Jan 20 '17 at 11:16

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.