Zero-variance boundary of an overflow constraint ($\mathbb{P}(X\ge C)=1_{\{m\ge C\}}$)

If a random variable has variance zero and mean $m$, it equals $m$ almost surely. For a target probability strictly between zero and one, $\mathbb{P}(X\ge C)$ meets the target exactly when $m<C$. Substituting zero variance into a positive-variance non-strict quantile constraint incorrectly includes $m=C$, where the overflow probability is one. The distinction comes from the atom at the threshold.