probability - Bounded by exponential implies uniformly integrable?

Suppose I have a sequence of r.v.'s $\left\{X_n\right\}_{n\in\mathbb{N}}$. I have shown that $\mathbb{P}\left( \left|X_n\right| > a\right) < C_1e^{-C_2 a}$ for all $n$. From here, I can show that $\lim_{a\to\infty} \sup_n \mathbb{P}\left(\left| X_n\right| > a\right) = 0$. My question is, does this imply that $X_n$'s are uniformly integrable?

I know that if $\sup_n \left|X_n\right| < Y$ and $\mathbb{E}\left[Y\right] < \infty$, then $X_n$'s are uniformly integrable. But do I have to construct an explicit random variable here? This notes says that

Answer

You don't have necessarily to construct such a random variable $Y$ (it is a sufficient condition, but not a necessary one). One possible approach to prove the uniform integrability is the following:

For any non-negative random variable $Z$ it holds that

$$\mathbb{E}(Z) = \int_0^{\infty} \mathbb{P}(Z > r) \, dr.$$

Applying this to $Z := 1_{|X_n|>a} |X_n|$ (for fixed $n$ and $a$) we find

$$\begin{align*} \int_{|X_n|>a} |X_n| \, d\mathbb{P} &= \int_0^a \mathbb{P}(|X_n|>a) \, dr +\int_a^{\infty} \mathbb{P}(|X_n|>r) \, dr \\ &\leq C_1a e^{-C_2a } + C_1 \int_a^{\infty} e^{-C_2r} \, dr. \end{align*}$$

Thus,

$$\sup_{n \geq 1} \int_{|X_n|>a} |X_n| \, d\mathbb{P} \leq C_1a e^{-C_2a } + C_1 \int_a^{\infty} e^{-C_2 r} \, dr \xrightarrow[]{a \to \infty} 0$$

which shows that $(X_n)_{n \geq 1}$ is uniformly integrable.