= Solution
For $k\ge1$, the <event> $\{\lfloor X\rfloor\ge k\}$ equals $\{X\ge k\}$. Integrating the <Pareto distribution>'s density gives
$$
\boxed{\mathbb P(\lfloor X\rfloor\ge k)=k^{1-t},\qquad
\mathbb P(\lfloor X\rfloor=k)=k^{1-t}-(k+1)^{1-t}\quad(k\ge1).}
$$
The mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in\mathbb N$, include the separate endpoint values \b[$\mathbb P(\lfloor X\rfloor\ge0)=1$ and $\mathbb P(\lfloor X\rfloor=0)=0$], since $X\ge1$ almost surely. Substituting $k=0$ into the power expression would be invalid.
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