Does $\frac{W_t}{\sqrt{t}}$ is uniformly integrable?

Question

My question is does $ \frac{W_t}{\sqrt{t}} $ is uniformly integrable?

Since $W_t \sim N(0,t) \Rightarrow \frac{W_t}{\sqrt{t}} \sim N(0,1)$, I tried to prove it this way:

$\mathbb{E}(|\frac{W_t}{\sqrt{t}}| 1_{\{|\frac{W_t}{\sqrt{t}}|>K\}}) = 2\int_{K}^{\infty}w\frac{e^{-w^2/2}}{\sqrt{2 \pi}}=\frac{2}{\sqrt{2 \pi}}e^{-K^2/2}$

But this expression $\frac{2}{\sqrt{2 \pi}}e^{-K^2/2}$ can be as small as we want.
Indeed for any given $\epsilon>0$ you can choose a $K>\sqrt{2ln(\sqrt{2/\pi})/\epsilon}$ that will verify: $\mathbb{E}(|\frac{W_t}{\sqrt{t}}| 1_{\{|\frac{W_t}{\sqrt{t}}|>K\}})< \epsilon$.
So it seems to me that the expression: $\frac{W_t}{\sqrt{t}}$ is UI, is it correct?

Thank.

Answer 1

Thank to Andrew Zhang. I copy past here his answer to mark this topic as solved.

"Looks good to me. Another way, any L1 random variable is uniformly integrable by dominated convergence."