I am learning maximum likelihood and UMVUE in statistical inference and I came across the indicator functions being used as follows: $$L(X, \theta) = \prod_{i=1}^n I_{ (\theta, \theta+1)} (x_i) = I_{ (\theta, \theta+1)} (x_n) I_{ (\theta, \theta+1)} (x_1) = I_{ (\ x_{(n)} - 1, \ inf)}(\theta) {I_{ (-inf , x_{(1)})}(\theta)} $$
Can someone help me understand the second and third steps here?Specifically how indicator function of x is converted to a function of theta
Definition: The function $I_{(a,b)}(x)$ is $1$ if $x \in (a,b)$ and $0$ elsewhere.
Your Problem: Start with $$ L(X, \theta) = \prod_{i=1}^n I_{ (\theta, \theta+1)} (x_i) $$
We'll show that this equals $I_{(\theta,\theta+1)}(x_{(1)}) I_{(\theta,\theta+1)} (x_{(n)})$. Clearly, $L$ takes on values $\{0,1\}$. If $L(X,\theta) =1$, then the result is trivial. If $L(X,\theta)=0$, then either $x_j > \theta+1$ or $x_j < \theta$ for some $j$ , which respectively imply $x_{(n)} > \theta+1$ or $x_{(1)}<\theta$ so that $I_{(\theta,\theta+1)}(x_{(1)}) I_{(\theta,\theta+1)} (x_{(n)})=0$.
So $L(X,\theta) = I_{(\theta,\theta+1)}(x_{(1)}) I_{(\theta,\theta+1)} (x_{(n)})$. The RHS equals $0$ besides the following intersection, where it is identically $1$: when $\theta < x_{(1)}< \theta+1$ and $\theta < x_{(n)} < \theta+1$; i.e. when $$ - \infty < \theta < x_{(1)} \le x_{(n)} < \theta + 1 < \infty;$$ i.e. when $-\infty < \theta < x_{(1)}$ and $x_{(n)}-1 < \theta < \infty$ . Thus $L(X,\theta)=I_{(x_{(n)} -1, \infty)}(\theta) I_{(-\infty,x_{(1)})}(\theta)$.