Rate function under scalar clipping ($K(a)={\inf }_{x\le a}I(x),K(b)={\inf }_{x\ge b}I(x)$)

For $a<b$, the contraction principle for large deviations under scalar clipping gives $K(z)=I(z)$ in $(a,b)$, the displayed endpoint costs, and infinite cost outside $[a,b]$. The endpoint costs can differ from $I(a)$ and $I(b)$ because clipping collects a whole input tail into a single output atom.