I'm studying uniform integrability of random variables and am confused by the difference between some definitions regarding $L^p$ spaces. Please let me know why the following is incorrect:
A random variable $X$ is in $L^p(\Omega,\mathcal{F},\mathbb{P})$ for $p\in[1,\infty)$ if $||X||_p = (\mathbb{E}[|X|^p])^{\frac{1}{p}}<\infty$. I'm not sure whether there should be a power of $\frac{1}{p}$ here - I've seen this defined without it. I will continue with $p=1$ as I am not sure whether this exponent is needed in definitions and the $p=1$ case is not affected by this:
$X$ is in $L^1$ if $\mathbb{E}[|X|] = \int_{\Omega}|X|d\mathbb{P}< \infty$. That is to say, $X \in L^1$ if $X$ is $\mathbb{P}$-integrable.
A collection of random variables $\Lambda$ is in $L^1(\Omega,\mathcal{F},\mathbb{P})$ $X \in L^1(\Omega,\mathcal{F},\mathbb{P})$ for all $X \in \Lambda$.
Now a collection random variables $\Lambda$ is $L^1$-bounded if $\sup_{X \in \Lambda} \mathbb{E}[|X|] < \infty$.
My question is: are all random variables or collections of random variables that are in $L^1$ not by definition $L^1$-bounded? A collection is in $L^1$ if all of its random variables are in $L^1$, but if this is the case $\mathbb{E}[|X|]<\infty$ for all $X \in \Lambda$. Then does it not follow that the supremum of these quantities is also bounded? Or is there nuance here that I am missing?
Thanks
**Edit - I have seen a theorem that says: If $\Lambda \subset L^1$ and is uniformly integrable, then $\Lambda$ is $L^1$-bounded. So I guess this answers my question since it implies that being in $L^1$ is not equivalent to being $L^1$ bounded. So could someone please clarify why that is the case? Thanks
EDIT: This is just what @RhysSteele said above, but I already typed up this answer before seeing his.
Consider $\Omega = [0,1]$ with the Borel $\sigma$-algebra $\mathcal{F}$ and the Lebesgue measure $\lambda$. Then $(\Omega,\mathcal{F},\lambda)$ is a probability space. On this space, define the sequence $(X_n)_{n\in\mathbb{N}}$ by $X_n = n$ for all $n\in\mathbb{N}$. Clearly, $$ \int_0^1 \vert X_n \vert \,\text{d}\lambda = n$$ for all $n\in\mathbb{N}$, i.e. each $X_n$ is integrable, hence $(X_n)_{n\in\mathbb{N}} \subset L^1(\Omega,\mathcal{F},\lambda)$. But clearly, $$ \sup_{n\in\mathbb{N}} \int_0^1 \vert X_n \vert \,\text{d}\lambda = \sup_{n\in\mathbb{N}} n = \infty,$$ which means that the sequence $(X_n)_{n\in\mathbb{N}}$ is not bounded in $L^1(\Omega,\mathcal{F},\lambda)$ even though each random variable is integrable.
Hope this answers your question.