uniform integrability does not imply $L_p$ boundednes

Question

Let $\mathcal H$ denote a family of random variables on ($\Omega,\mathcal A ,P$) . We all know that If $\mathcal H$ be $L_p$ bounded then $L_p$ is uniformly integrable but the converse is not true. How can I give an example for this and how can I show that this is not uniformly integrable.

Answer 1

Let $X \in L_1$ be an integrable random variable that is not $L_p$ integrable, and set $\mathcal H := \{X\}$. Then $\mathcal H$ is uniformly integrable because it is a singleton, but is not $L_p$ bounded because ${\sup_{Y \in \mathcal H} \|Y\|_p = \|X\|_p = \infty.}$

Answer 2

In general:
$L^p$ bounded $\Longrightarrow$ uniformly integrable $\Longrightarrow$ $L^1$ bounded.
So if your question is for a fixed $p$, Then we can choose $s$ with $1 < s < p$ and find a family $\mathcal H$ that is $L^s$ bounded but not $L^p$ bounded. Then $\mathcal H$ will be uniformly integrable. On the probability space $[0,1]$, such an example can be found of the form $\{f_n : n = 1,2,3,\cdots\}$ where $f_n(x) = \min\{n,x^\alpha\}$ for an appropriate choice of $\alpha$. Choose $\alpha$ so that $$ \int_0^1x^{\alpha p}\;dx = +\infty\quad\text{but}\quad\int_0^1x^{\alpha s}\;dx < +\infty $$

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We can insert another space here. Define $\mathcal H$ is $L\log L$ bounded iff $\sup_{X \in \mathcal H}\mathbb E[|X| \log^+|X|] < +\infty$. Where $\log^+(t) = \max\{0,\log t\}$. Then
$L^p$ bounded $\Longrightarrow$ $L\log L$ bounded $\Longrightarrow$ uniformly integrable.
So for a more powerful counterexample, choose $\mathcal H$ to be $L\log L$ bounded, but $L_p$ unbounded for all $p>1$. On the space $[0,1/2]$, such an example can be found of the form $\{f_n : n = 1,2,3,\cdots\}$ where $f_n(x) = \min\{n,x^\alpha\;|\log(x)|^\beta\}$ for an appropriate choice of $\alpha, \beta$.
Choose $\alpha, \beta$ so that $$ \int_0^{1/2}|x^{\alpha}\log^\beta(x)|^p\;dx = +\infty\quad \text{ for all } p>1\\ \qquad\text{but}\quad \int_0^{1/2}|x^\alpha\log^\beta(x)|\log^+|x^\alpha\log^\beta(x)|\;dx< +\infty $$ A choice that works is $\alpha=-1, \beta=-3$.