Let $g(x)$ be a positive integrable function on $(0,\infty)$. Suppose $X_1,\ldots,X_n$ are random independent variables with common density $$f(x) = \dfrac{g(x)}{\int_\theta^\infty g(x)dx}\mathbb{1}_{[\theta,\infty)}(x)$$
Exercise: Show that $T(X) = \min(X_1,\ldots,X_n)$ is a sufficient statistic using the factorisation theorem.
What I've tried: To show that $T(X)$ is sufficient I need to write $f(X\mid\theta)$ in the form of the factorisation theorem $$f(x\mid\theta) = g_\theta(T(X))h(x)$$
I have that $f(x\mid\theta) = \prod_{i = 1}^n\dfrac{g(x)}{\int_\theta^\infty g(x)dx}\mathbb{1}_{[\theta,\infty)}(x)$. If $X_{(1)} \geq \theta$, then every $X_i \geq \theta$, so we can replace the indicator function so that it only depends on $X_{(1)}$, we get $f(x\mid\theta) = \prod_{i = 1}^n\dfrac{g(x)}{\int_\theta^\infty g(x)dx}\mathbb{1}_{\{X_{(1)}\geq\theta\}}$. Unfortunately, I don't see how I can make the $\int_\theta^\infty$ part a function of $X_{(1)}$ alone. I have $h(x) = \prod_{i = 1}^n g(x)$, but I can't find the right way to formulate $g(X)$.
Question: How do I solve this exercise, as indicated, using the factorisation theorem?
Thanks
Note that $$f(x_i;\theta) = \left\{ \begin{matrix}\frac{g(x_i)}{\int_{\theta}^{\infty} g(t)dt} & \text{if $x \ge \theta$;}\\0 & \text{if $ x_i < \theta $.} \end{matrix}\right.$$ Consequently, $$f(x_i;\theta) = \frac{g(x_i)}{l(\theta)} 1_{\{x_i\ge \theta \}},$$ where $l(\theta)=\int_{\theta}^{\infty} g(t)dt$ is a function of $\theta$ only (i.e. it does not depend on $x_i$). Therefore, $$f(x_1,x_2,\cdots, x_n;\theta) = \color{red}{\left(\prod_{i=1}^{n}g(x_i)\right)} \color{blue}{\frac{1}{{l(\theta)}^n}\prod_{i=1}^{n} 1_{\{x_i\ge \theta \}}}.$$ Since $$\prod_{i=1}^{n} 1_{\{x_i \ge \theta\}} = 1_{\min\{x_i\}\ge \theta},$$
$$f(\theta|x_1,x_2,\cdots, x_n) = \color{red}{h(x_1,x_2,\cdots,x_n)}\color{blue}{g_{\theta}(\min\{x_i\})},$$ where $$\color{red}{h(x_1,x_2,\cdots,x_n) = \prod_{i=1}^{n}g(x_i)},$$ and $$\color{blue}{g_\theta(T(X)) = \frac{1}{{l(\theta)}^n}1_{\{T(X) \ge \theta\}}}.$$