Are there any grounds for the following claim?
For distributions for which the support in $x$ depends on $\theta$, the CRLB $=0$.
For a quick example (Casella-Berger, Example 7.3.13): Let $X_1,\dots,X_n$ be iid with pdf $f(x\mid\theta) = 1/\theta, 0 < x <\theta$. There, CB used the typical formula for Fisher information $I(\theta)$ to get CRLB $= \theta^2/n$ and subsequently showed that for the unbiased estimator $\hat{\theta} = \frac{n+1}{n}X_{(n)}$, $\mathrm{Var}\hat{\theta} = \frac{1}{n(n+2)}\theta^2$ which is uniformly smaller than the CRLB and thus Cramer-Rao Inequality is violated.
However, my lecturer disagrees with this approach as CB ignored the indicator function $\mathbf{1}_{(0<x<\theta)}$ when taking the derivative of the log-likelihood function. His approach was to note that because the likelihood function is not continuous in $\theta$, it is not differentiable in $\theta$. In situations like these, we should define $I(\theta) = +\infty$ and thus CRLB $= 0$, in which case the Cramer-Rao Inequality was not violated.
I Google'd a bit but did not find any references to validate this alternative approach although it does make some sense (although I can't seem to grasp the significance of having CRLB $= 0$). Has anyone come across something like this before?