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The jump activity of a Lévy process describes whether it has finitely or infinitely many jumps in a bounded time interval. Its Lévy measure $\nu$ decides this: finite activity corresponds to $\nu ({\mathbb{R}}^d\setminus \{0\})<\infty$, in which case the jump component is a Compound Poisson process. Infinite mass gives infinitely many small jumps in every positive-length time interval, almost surely. For every $\epsilon >0$, however, jumps larger than $\epsilon$ still have a finite rate.

Activity and total variation are different. With zero Brownian component, sample paths have finite variation on compact time intervals exactly whenFor example, $\nu (dx)=c{x}^{-1-\alpha }dx$ on $(0,\infty )$, with $c>0$ and $0<\alpha <1$, has infinite total mass but a finite integral of $x$ near zero. It therefore permits infinite jump activity with finite variation. Conversely, Brownian motion has no jumps but infinite total variation on every positive-length time interval, almost surely.^[ref]

For $\alpha >0$, the Lévy process with characteristic function $\mathbb{E}{e}^{iu{X}_t}=(1+u^2/{\alpha }^2{)}^{-t}$ has zero Gaussian coefficient, zero uncompensated drift, and the displayed Lévy measure. Indeed, the symmetric jump integral is $2{\int }_0^{\infty }(\cos (uy)-1)e^{-\alpha y}dy/y$. Differentiating in $u$ gives $-2u/({\alpha }^2+u^2)$, and integrating back from zero gives $-\log (1+u^2/{\alpha }^2)$. The time-one law is the Laplace distribution of rate $\alpha$. The jump activity is infinite, since $K$ has infinite mass near zero, but the jump variation is finite because $\int (1\wedge |y|)K(dy)<\infty$.