$\require{cancel}$
John Xmith published an interesting paper [1] in which he reported the following result, which, unfortunately, is mistaken: $$ \xcancel{\begin{align} & \textbf{Theorem 2: } \text{Let $X$ be } \cdots\cdots\cdots\cdot \\ & \text{and further assume blah blah blah blah} \\ & \text{Then} \\ & \qquad \qquad \int\cdots \cdots = \sum \cdots\cdots. \end{align}} $$ Here, we will exhibit a counterexample to this erroneous proposition.