Let $\mathcal{Y}$ be a collection of random variables on a probability space $(\Omega,\mathcal{F},P)$ that is bounded in $L^p$-norm, $p \in (1,\infty)$. Fix $X \in L^q$ where $L^q$ is the dual space of $L^p$.
Is $X\mathcal{Y} := \{XY : Y \in \mathcal{Y} \}$ also uniformly integrable?
I think the answer is yes, but I cannot seem to prove it. Of course $\mathcal{Y}$ is uniformly integrable. Any help is appreciated. I was thinking of making use of Young's inequality: $ab \leq a^p/p + b^q/q$, for $a,b \geq 0$. Thus \begin{align*} \mathbb{E}[|XY| \mathbf{1}_{|XY| > k} ] &\leq \mathbb{E}[|XY| \mathbf{1}_{|X|^q/q+|Y|^p/p > k} ] \\ &\leq \mathbb{E}[|XY| (\mathbf{1}_{|X|^q/q> k/2} + \mathbf{1}_{|Y|^p/p> k/2}) ] \\ &\leq \lVert X \rVert_{q}\lVert Y\mathbf{1}_{|Y|^p/p> k/2} \rVert_{p} + \lVert Y \rVert_{p}\lVert X\mathbf{1}_{|X|^q/q> k/2} \rVert_{q}. \end{align*}
Hence, $\lim_{k \to \infty} \sup_{Y \in \mathcal{Y}} \mathbb{E}[|XY| \mathbf{1}_{|XY| > k} ] = 0+0$?
The idea of your approach is fine, but your estimate for the term $\mathbb{E}(|XY| 1_{|Y|^p/p>k/2})$ is too rough (see the above comment). Here is a possible way to deal with it:
Fix $R>0$, then
\begin{align*} \mathbb{E}(|XY| 1_{|Y|^p/p>k/2}) &\leq R \mathbb{E}(|Y| 1_{|Y|^p/p>k/2}) + \mathbb{E}(|XY| 1_{|X|>R}) \\ &\leq R \mathbb{E}(|Y| 1_{|Y|^p/p>k/2}) + \|X 1_{|X|>R}\|_q \|Y\|_p. \end{align*}
Since $Y \in \mathcal{Y}$ is uniformly integrable and bounded in $L^p$, we have
$$\sup_{Y \in \mathcal{Y}} \mathbb{E}(|XY| 1_{|Y|^p/p>k/2}) \leq R \underbrace{\sup_{Y \in \mathcal{Y}} \mathbb{E}(|Y| 1_{|Y|^p/p>k/2})}_{\to 0 \, \text{as $k \to \infty$}} + \|X 1_{|X|>R}\|_q \underbrace{\sup_{Y \in \mathcal{Y}} \|Y\|_p}_{=:M<\infty},$$
i.e. $$\lim_{k \to \infty} \sup_{Y \in \mathcal{Y}} \mathbb{E}(|XY| 1_{|Y|^p/p>k/2}) \leq M \|X 1_{|X|>R}\|_q \xrightarrow[R \to \infty]{} 0.$$