Let $\Theta$ be a nonempty compact subset of $\mathbb R^d$. For example, the reader may think of the closed unit-ball $\Theta := \{\theta \in \mathbb R^d \mid \|\theta\|_2 \le 1\}$. Consider an estimator $\hat{T}:\mathcal X^* \to \Theta$, $(x_1,\ldots,x_n) \to \hat{\theta}_n :=\hat{T}(x_1,\ldots,x_n)$, where $\mathcal X^* = \cup_{n=1}^\infty \mathcal X^n$ is the set of all finite-tuples of points in $\mathcal X$, i.e datasets over $\mathcal X$.
Question. What is a reasonable way to define the breakdown point of $\hat{T}$ (i.e the fraction of data which can be contaminated without completely "screwing up" the output $\hat{\theta}_n$ of $\hat{T}$ on a dataset $x_1^n:=(x_1,\ldots,x_n)$ with $n \to \infty$). ?
Note. The main difficulty here is the fact that $\Theta$ is compact and so $\hat{T}$ is bounded.
Classical theory and the issue with bounded statistics
The breakdown point of an estimator $\hat{T}$ on a dataset $x_1^n$ is usually defined by
$$ \mathrm{BDP}(\hat{T},x_1^n) := \min_{1 \le m \le n}\left\{\frac{m}{n+m} \mid \sup_{y_1^m \in \mathcal X^m}\|\Delta \hat{T}(x_1^n,y_1^m)\| = \infty\right\}, $$ where $\Delta \hat{T}(x_1^n,y_1^m):= \hat T(x_1^n,y_1^m)) - \hat T(x_1^n)$ and we use the convention that $\mathrm{BD}(\hat{T},x_1^n) = 1/2$ if the set above is empty.
However, the issue with this definition is the following: in my example above, since $\hat{T}$ is bounded (because $\Theta$ is compact), it always holds that $\mathrm{BDP}(\hat{T},x_1^n) = 1/2$, which is misleading, as it says that every estimator for a bounded statistics is optimally robust!
