 Solution

Let $m=\lfloor \alpha n\rfloor$. If at most $m$ observations are replaced arbitrarily, every order statistic retained between ranks $m+1$ and $n-m$ remains between the minimum and maximum of the unreplaced observations. The trimmed mean is therefore bounded as the replacement values diverge.

If more than $m$ observations are replaced by a common value tending to $+\infty$, at least one replacement remains after the largest $m$ observations are trimmed, and the trimmed mean tends to $+\infty$. The analogous construction tends to $-\infty$. Thus the largest fraction of arbitrary replacements for which boundedness is guaranteed isThis is the convention for the finite-sample replacement breakdown point used in the question; the alternative convention based on the smallest breaking fraction reports $(m+1)/n$.