= Solution
The support restricts $\theta$ to $\theta>0$, $\theta\leq X_{(1)}$ and $\theta\geq X_{(n)}-1$. Within its feasible interior, the <likelihood function> is proportional to $\prod_i(X_i-\theta)^{-\alpha}$ and is strictly increasing in $\theta$, diverging as $\theta\uparrow X_{(1)}$. Under the printed infinite-endpoint-density, or extended-likelihood convention, the <maximum-likelihood estimator> is
$$
\boxed{\widehat\theta_n=X_{(1)}.}
$$
A density is defined only almost everywhere. If its value at the singular endpoint is reassigned a finite value, a finite maximum need not exist; the sample-minimum answer uses the stated endpoint/supremum convention.
Independence and $\mathbb P(Y\leq t)=t^r$ on $[0,1]$ give
$$
\boxed{\mathbb P(\widehat\theta_n-\theta>t)=\begin{cases}1,&t\leq0,\\(1-t^r)^n,&0<t<1,\\0,&t\geq1.\end{cases}}
$$
The error is positive almost surely, so the estimator is not an <unbiased estimator>. More explicitly its <bias> is
$$
\mathbb E(\widehat\theta_n-\theta)=\int_0^1(1-t^r)^n\,dt
=\frac{\Gamma(1+1/r)\Gamma(n+1)}{\Gamma(n+1+1/r)}>0.
$$
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