Let $((0,1), \mathcal{B}(0,1), \mathcal{Leb}_{(0,1)})$ be a probability space. Let $(X_n)_{n=1}^{\infty}$ be a sequence of random variables such that $X_n=\alpha^n\mathbb{I}_{(0,2^{-n})}$, $n \geq 1$, $\alpha \in(0,1)$.
Is $(X_n)_{n=1}^{\infty}$ uniformly integrable?
My attempt
We need to show that $(X_n)_{n=1}^{\infty}\subseteq{L^1}$ and that $k(b)=\sup_{n\in\mathbb{N}}\alpha^n\mathbb{I}_{(0,2^{-n})}\mathbb{I}_{\alpha^n\mathbb{I}_{(0,2^{-n})}\gt{b}} \underset{b \rightarrow \infty}{{\longrightarrow}}0$.
$\mathbb{E}\big |\alpha^n\mathbb{I}_{(0,2^{-n})}\big |=\big(\frac\alpha2\big)^n \lt\infty\ \forall \ n\geq1$, so $(X_n)_{n=1}^{\infty}\subseteq{L^1}$. Then (and this is where I am most uncertain)
$\mathbb{E}\ \alpha^n\mathbb{I}_{(0,2^{-n})}\mathbb{I}_{\alpha^n\mathbb{I}_{(0,2^{-n})}\gt{b}} = \begin{cases} \big(\frac\alpha2\big)^n, \ \ \ \ \alpha^n \gt b \\ 0\ \ \ \ \ \ \ , \ \ \ \ \alpha^n \leq b \end{cases} $
which is decreasing in $n$, so it should be
$k(b)=\sup_{n \in \mathbb{N}}\mathbb{E}\ \alpha^n\mathbb{I}_{(0,2^{-n})}\mathbb{I}_{\alpha^n\mathbb{I}_{(0,2^{-n})}\gt{b}}=\frac\alpha2\mathbb{I}_{(0,\alpha)}(b) \underset{b \rightarrow \infty}{{\longrightarrow}}0$.
I am not sure what I did is correct. Can anyone confirm or correct this? Any help is much appreciated.
Uniform integrability follows directly from $$ \lvert X_n\rvert= \alpha^n\mathbb{I}_{(0,2^{-n})}\leqslant \alpha^n\mathbb{I}_{(0,1)}\leqslant \mathbb{I}_{(0,1)}. $$