# Why $\mathbb E \left[ \exp (u X_t )\right] < \infty$ if and only if $\int_{|z|>1} e^{uz} ~\nu(dz) < \infty$ for a Lévy process?

Consider a Lévy process $\left( X_t\right)_{t\geq 0}$ whose Lévy measure is $\nu$.

Why have we for all $u \in \mathbb R$ the following result?

$\mathbb E \left[ \exp (u X_t )\right] < \infty$ if and only if $\int_{|z|>1} e^{uz} ~\nu(dz) < \infty$

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Call (1) the condition that $\mathbb E(\mathrm e^{uX_t})$ is finite for every nonnegative $t$ and $u$, and (2) the condition that $\displaystyle\int_{x\geqslant1}\mathrm e^{ux}\nu(\mathrm dx)$ is finite for every nonnegative $u$.
Likewise, $\mathbb E(\mathrm e^{-uX_t})$ is finite for every nonnegative $t$ and $u$ if and only if $\displaystyle\int_{x\leqslant-1}\mathrm e^{-ux}\nu(\mathrm dx)$ is finite for every nonnegative $u$.