Why do we assume all sample $Y\gt 0$ in uniform distribution of $[0, \theta]$?

Question

So I am calculating sufficient statistic for uniform distribution on $[0,\theta]$ and $\theta \gt 0$. Sample $Y=(Y_1,Y_2,...,Y_n)$ has size $n$. I have $L(\theta, Y)=\prod^n_{i=1}\frac{1}{\theta-0} \mathbb{1}(0<Y<\theta)= \frac{1}{{\theta}^n} \mathbb{1}(\min(Y)>0)*\mathbb{1}(\max(Y)<\theta)$. I looked up answer online and discovered we can simply assume $\min{Y}>0$ and take $\max(Y)$ as the sufficient statistics. I know this problem can be solved by using Factorization theorem and taking $g(T(Y), \theta)=\frac{1}{{\theta}^n}\mathbb{1}(\max(Y)<\theta)$ and $h(Y)=\mathbb{1}(\min(Y)>0)$. My confusion is at why can we assume $\min{Y}>0$, I think sample data can certainly be below $0$.